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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 7
The value of $$C_1 ^2+C_2 ^2....+C_n ^2$$ (where $$C_i$$ is the $$i^{th}$$ coefficient of $$(1+x)^n$$ expansion), is:
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$$\dfrac {n^n}{n!}$$
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$$\dfrac {2n!}{n!n!}$$
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$$\dfrac {2n!}{n!}$$
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$$\dfrac {n!\times2^n}{2n!}$$
Explanation
$${ \left( 1+x \right) }^{ n }=^{ n }{ { C }_{ 0 } }+x\left( ^{ n }{ { C }_{ 1 } } \right) { +{ x }^{ 2 }\left( ^{ n }{ { C }_{ 2 } } \right) + }{ x }^{ 3 }\left( ^{ n }{ { C }_{ 3 } } \right) +...+{ x }^{ n }\left( ^{ n }{ { C }_{ n } } \right) \\ $$
$${ \left( x+1 \right) }^{ n }={ x }^{ n }\left( ^{ n }{ { C }_{ 0 } } \right) +{ x }^{ n-1 }\left( ^{ n }{ { C }_{ 1 } } \right) +...+{ x }^{ 0 }\left( ^{ n }{ { C }_{ n } } \right) $$
Now, $$\quad \sum _{ r=0 }^{ n }{ { \left( ^{ n }{ { C }_{ r } } \right) }^{ 2 } } $$= co-efficient of $$x^n$$ in $$(1+x)^{2n}=^{ 2n }{ { C }_{ n } }$$
Thus, $$^{ 2n }{ { C }_{ n } }={ \left( ^{ n }{ { C }_{ 0 } } \right) }^{ 2 }+{ \left( ^{ n }{ { C }_{ 1 } } \right) }^{ 2 }+...{ \left( ^{ n }{ { C }_{ n } } \right) }^{ 2 }$$, where $${ \left( ^{ n }{ { C }_{ i } } \right) }$$ is the $$i^{th}$$ coefficient of $$(1+x)^n$$ expansion.
So, $$C_1^2+C_2^2+...+C_n^2={ \left( ^{ 2n }{ { C }_{ n } } \right) }=\dfrac { 2n! }{ n!n! } $$
If $$f(n)=\sum_{s=1}^n \sum_{r=s}^n \:^nC_r \:^rC_s$$, then $$f(3)=$$
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$$27$$
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$$19$$
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$$1$$
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$$5$$
Explanation
Given : $$f\left( n \right) =\overset { n }{ \underset { s=1 }{ \Sigma } }\; \overset { n }{ \underset { r=s }{ \Sigma } } { ^{ n }{ C } }_{ r }{ ^{ r }{ C } }_{ s }$$
To Find : $$f\left( 3 \right) =?$$
Sol. : $$f\left( n \right) =\left\{ { ^{ n }{ C } }_{ 1 }{ ^{ 1 }{ C } }_{ 1 }+{ ^{ n }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 1 }+{ ^{ n }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 1 }......{ ^{ n }{ C } }_{ n }.{ ^{ n }{ C } }_{ 1 } \right\} +\left\{ { ^{ n }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 2 }+{ ^{ n }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 2 }+.....{ ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ 2 } \right\} +.....$$
$$+\left\{ { ^{ n }{ C } }_{ n-1 }{ ^{ n-1 }{ C } }_{ n-1 }+{ ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n-1 } \right\} +\left\{ { ^{ n }{ C } }_{ n }{ ^{ n }{ C } }_{ n } \right\} $$
$$\Rightarrow$$ Take $$n=3$$
$$\Rightarrow f\left( 3 \right) =\left\{ { ^{ 3 }{ C } }_{ 1 }{ ^{ 1 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 1 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 1 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 2 }{ ^{ 2 }{ C } }_{ 2 }+{ ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 2 } \right\} +\left\{ { ^{ 3 }{ C } }_{ 3 }{ ^{ 3 }{ C } }_{ 3 } \right\} $$
$$=\left\{ 3+3.2+3 \right\} +\left\{ 3+3 \right\} +\left\{ 1 \right\} $$
$$=19$$
Hence, the answer is $$19.$$
The value of $$(n+2)C_02^{n+1}-(n+1)C_12^n+nC_22^{n-1}+....$$ is equal to:
$$(C_r=\:^nC_r)$$
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$$4n$$
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$$4$$
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$$2n+4$$
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$$4+4n$$
Explanation
We know, $${ \left( 1-x \right) }^{ n }=_{ }^{ n }{ { C }_{ 0 }{ x }^{ n } }-_{ }^{ n }{ { C }_{ 1 }{ x }^{ n-1 } }+_{ }^{ n }{ { C }_{ 2 }{ x }^{ n-2 } }-...._{ }^{ n }{ { C }_{ n } }$$
Multiplying by $$x^2$$ on both sides,
$${ x }^{ 2 }{ \left( 1-x \right) }^{ n }=_{ }^{ n }{ { C }_{ 0 }{ x }^{ n+2 } }-_{ }^{ n }{ { C }_{ 1 }{ x }^{ n+1 } }+_{ }^{ n }{ { C }_{ 2 }{ x }^{ n } }-....\\ $$
Taking the derivative,
$$2x{ \left( 1-x \right) }^{ n }+n{ x }^{ 2 }{ \left( 1-x \right) }^{ n-1 }=\left( n+2 \right) { { C }_{ 0 }{ x }^{ n+1 } }-\left( n+1 \right) { { C }_{ 1 }{ x }^{ n } }+n{ { C }_{ 2 }{ x }^{ n-1 } }-....$$
Putting $$x=2$$ in LHS, we get our required result,
$$\quad4(-1)^n+n(4)(-1)^{n-1}$$
$$=4-4n$$, $$n$$ is even
$$=4n-4$$, $$n$$ is odd.
There is $$no$$ $$option$$ matching the answer.
If $$^{n-1}C_{r}=(k^2-3)\:^nC_{r+1}$$, then $$k\:\:\epsilon$$
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$$(-\infty,-2]$$
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$$[2,\infty)$$
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$$[-\sqrt 3,\sqrt 3]$$
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$$(\sqrt 3,2]$$
Explanation
Given, $$^{n-1}{C}_{r}=(k^2-3)^{n}{C}_{r+1}$$
$$\Rightarrow$$$$\dfrac{^{n-1}{C}_{r}}{^{n}{C}_{r+1}}=(k^2-3)$$ ....Using $$^{n}{C}_{x}=\dfrac{n}{x}.^{n-1}{C}_{x-1}$$
$$\Rightarrow$$$$\dfrac{r+1}{n}=(k^2-3)$$ ....As $$n>r+1$$
So, $$0<\dfrac{r+1}{n}\le1$$
Therefore, $$0<k^2-3\le1$$
$$\Rightarrow 3<k^2\le4$$
$$\Rightarrow \sqrt{3}<k\le2$$
So, k$$\in$$$$(\sqrt{3},2]$$
If $$P_n$$ denotes the product of all the coefficients in the expansion of $$(1+x)^n$$, then $$\dfrac {P_{n+1}}{P_n}$$ is equal to:
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$$\dfrac {(n+2)^n}{n!}$$
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$$\dfrac {(n+1)^{n+1}}{n+1!}$$
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$$\dfrac {(n+1)^{n+1}}{n!}$$
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$$\dfrac {(n+1)^n}{n+1!}$$
Explanation
$$(1+x)^n$$
$${ P }_{ n+1 }=_{ }^{ n+1 }{ { C }_{ 0 } }\times _{ }^{ n+1 }{ { C }_{ 1 } }\times _{ }^{ n+1 }{ { C }_{ 2 } }\times ........\times _{ }^{ n+1 }{ { C }_{ n+1 } }\\ { P }_{ n }=_{ }^{ n }{ { C }_{ 0 } }\times _{ }^{ n }{ { C }_{ 1 } }\times _{ }^{ n }{ { C }_{ 2 } }\times ........\times _{ }^{ n }{ { C }_{ n } }$$
$$\quad \prod _{ r=0 }^{ n }{ \dfrac { _{ }^{ n+1 }{ { C }_{ r } } }{ _{ }^{ n }{ { C }_{ r } } } \quad \quad \quad \quad \quad \quad \quad \left( \because _{ }^{ n+1 }{ { C }_{ n+1 }=1 } \right) } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1)! }{ r!(n+1-r)! } } \times \dfrac { r!(n-r)! }{ n! } \\ =\prod _{ r=0 }^{ n }{ \dfrac { (n+1) }{ \left( n-r+1 \right) } } \\ ={ \left( n+1 \right) }^{ n }\prod _{ r=0 }^{ n }{ \dfrac { 1 }{ n-r+1 } } \\ =\dfrac { { \left( n+1 \right) }^{ n } }{ \left( n+1 \right) ! } $$
Hence, the correct option is $$D$$.
The value of $$\displaystyle \:^{50}C_4+\sum_{r=1}^6 \:^{56-r}C_3$$ is
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$$\:^{55}C_4$$
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$$\:^{55}C_3$$
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$$\:^{56}C_3$$
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$$\:^{56}C_4$$
Explanation
$$^{50}{C}_{4}$$+ $$\sum _{ r=1 }^{6 }{^{56-r}{C}_{3} }$$
$$^{50}{C}_{4}$$+$$^{50}{C}_{3}$$+$$^{51}{C}_{3}$$+.................+$$^{55}{C}_{3}$$
Using $$^{n}{C}_{r}$$+$$^{n}{C}_{r-1}$$=$$^{n+1}{C}{r}$$
$$^{51}{C}_{4}$$+$$^{51}{C}_{3}$$+.................+$$^{55}{C}_{3}$$
Similarly the series would become like this
$$^{55}{C}_{4}$$+$$^{55}{C}_{3}$$
Using $$^{n}{C}_{r}$$+$$^{n}{C}_{r-1}$$=$$^{n+1}{C}{r}$$
$$^{56}{C}_{4}$$
The coefficient of $$x^{53}$$ in the following expansions.
$$\displaystyle \sum_{m=0}^{100} \,^{100}C_m(x-3)^{100-m}\cdot 2^m$$ is
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$$^{100}C _{47}$$
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$$^{100}C _{53}$$
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$$^{-100}C _{53}$$
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$$^{-100}C _{100}$$
Explanation
$$\displaystyle \sum_{m=0}^{100} \, ^{100}C_m(x-3)^{100-m}\cdot 2^m$$
Above expansion can be rewritten as
$$[(x-3)+2]^{100}=(x-1)^{100}=(1-x)^{100}$$
$$\therefore x^{53}$$ will occur in $$T_{54}$$ i.e. $$54^{th}$$ term.
So, $$m=53$$
$$\Rightarrow T_{54}={ }^{100}C_{53}(-x)^{53}$$
$$\therefore $$ Required coefficient is $${ }^{-100} C_{53}$$.
The coefficient of $$x^{2012}$$ in $$\dfrac{1+x}{(1+x^2)(1-x)}$$ is.
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1
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2
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3
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4
Explanation
$$\displaystyle{\frac { (1+x) }{ (1+{ x }^{ 2 })(x-1) } =\frac { x }{ (1+{ x }^{ 2 }) } +\frac { 1 }{ (1-x) } }$$
The general term for $$\displaystyle{{ \left( 1-x \right) }^{ -1 }=1+x+{ x }^{ 2 }+{ x }^{ 3 }+.........}$$
$$\displaystyle{{ \left( 1-x \right) }^{ -1 }=\sum _{ 0 }^{ \infty }{ { x }^{ n } } .........(i)}$$
Replacing $$x$$ by $$-{ x }^{ 2 }$$
$$\displaystyle{\frac { 1 }{ 1+{ x }^{ 2 } } =1-{ x }^{ 2 }+{ x }^{ 4 }-{ x }^{ 6 }.......\\ \frac { 1 }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n } }} $$
Multiplying both sides by $$x$$
$$\displaystyle{\frac { x }{ 1+{ x }^{ 2 } } =\sum _{ 0 }^{ \infty }{ { { (-1) }^{ n }x }^{ 2n+1 } } ....(ii)}$$
From (i) clearly the cofficient of $${ x }^{ 2012 }$$ is $$1$$ and there is no term with even cofficient in $$(ii)$$
So the cofficient of $$\displaystyle{{ x }^{ 2012 }}$$ in the given expression is $$1$$
Let $$n$$ be a positive integer and, $${ \left( 1+x \right) }^{ n }={ a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+\dots +{ a }_{ n }{ x }^{ n }$$. What is $${ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }+\dots +{ a }_{ n }$$ equal to?
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$$1$$
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$${ 2 }^{ n }$$
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$${ 2 }^{ n-1 }$$
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$${ 2 }^{ n+1 }$$
Explanation
Given expansion is $${(1+x)}^n=a_0+a_1 x+a_2 x^2+....a_n x^n$$
Substituting $$x=1$$ on both the sides we get,
$${(1+1)}^n=a_0+a_1\times 1+a_2\times 1^2+....a_n\times 1^n$$
$$\implies 2^n=a_0+a_1+a_2+...a_n$$
Thus, the value of $$a_0+a_1+a_2+...a_n$$ is $$2^n$$.
$$5^{th}$$ term from the end in the expansion of $$\left( \dfrac{x^2}{2} - \dfrac{2}{x^2} \right )^{12}$$ is
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$$-7920x^{-4}$$
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$$7920x^{4}$$
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$$7920x^{-4}$$
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$$-7920x^{4}$$
Explanation
$$ { 5 }^{ th }$$ term in the expansion of$${ \left( \cfrac { { x }^{ 2 } }{ 2 } -\cfrac { 2 }{ { x }^{ 2 } } \right) }^{ 12 }$$ is
$$ { T }_{ r+1 }= { }^nC_r { x }^{ r }.{ y }^{n- r }$$
$$ { T }_{ 5 }= { }^{12}C_4 \left[{ \left( { \cfrac { { x }^{ 2 } }{ 2 } } \right) ^{ 4 }{ \left( \cfrac { -2 }{ { x }^{ 2 } } \right) }^{ 8 } }\right]$$
$$ \quad \quad = \cfrac { 12! }{ 4!8! } .\cfrac { { x }^{ 8 } }{ { 2 }^{ 4 } } .\cfrac { { 2 }^{ 8 } }{ { x }^{ 16 } }$$
$$ \quad \quad = { 7920x }^{ -4 } $$
Consider the expansion of $$(1+x)^{2n+1}$$
The average of the coefficients of the two middle terms in the expansion is
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$$^{2n+1}C_{n+2}$$
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$$^{2n+1}C_{n}$$
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$$^{2n+1}C_{n-1}$$
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$$^{2n}C_{n+1}$$
Explanation
Since, $$2n+1$$ is odd.
Hence, $$\cfrac{2n+1+1}{2}$$ and $$\cfrac{2n+1+3}{2}$$ are two middle terms.
i.e. $$(n+1)th$$ and $$(n+2)th$$ terms are two middle terms.
$$\therefore\cfrac{C^{2n+1}_n+ C^{2n+1}_{n+1}}{2}=\cfrac{C^{2n+1+1}_{n+1}}{2}$$
$$=\cfrac12C^{2n+2}_{n+1}=\cfrac12.\cfrac{2n+2}{n+1}C^{2n+1}_{n}=C^{2n+1}_{n}$$
Hence, B is the correct option.
In the expansion of $$\left( x^3 - \dfrac{1}{x^2} \right )^n , n \in N$$, if the sum of the coefficient of $$x^5$$ and $$x^{10}$$ is $$0$$, then $$n$$ is :
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$$25$$
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$$20$$
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$$15$$
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None of these
Explanation
Term of $${ x }^{ 5 }=^{ n }{ C }_{ r }\quad { x }^{ 3r }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }=^{ n }{ C }_{ r }\quad { x }^{ 3r }. { x }^{ 2r-2n }. { \left( -1 \right) }^{ n-r }$$
So $$3r+2r-2n=5$$
$$r=\cfrac { 5+2n }{ 5 } \longrightarrow \left( 1 \right) $$
Also for $${ x }^{ 10 }$$ be $$\left( { r }^{ 1 }+1 \right) ^{ th }$$ term
So term with $${ x }^{ 10 }=^{ n }C_{ { r }^{ 1 } }{ x }^{ { 3r }^{ 1 } }{ \left( \cfrac { -1 }{ { x }^{ 2 } } \right) }^{ n-r }$$
$$\Rightarrow 3{ r }^{ 1 }+2{ r }^{ 1 }-2n=10$$
$${ r }^{ 1 }=\cfrac { 2n+10 }{ 5 } \longrightarrow \left( 2 \right) $$
Now we know if $$\left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right| \Rightarrow n=r+{ r }^{ 1 }$$
Now we add co oefficient of $${ x }^{ 5 }\& { x }^{ 10 }$$
$$^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }+^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }=0$$
Co oefficient of $${ x }^{ 5 }\quad is\quad ^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }$$; co oeffiecient of $${ x }^{ 10 }\quad is\quad ^{ n }{ C }_{ { r }^{ 1 } }{ \left( -1 \right) }^{ { n-r }^{ 1 } }$$
$$\Rightarrow ^{ n }{ C }_{ r }{ \left( -1 \right) }^{ n-r }=-\left( ^{ n }{ C }_{ { r }^{ 1 } }\quad { \left( -1 \right) }^{ { n-r }^{ 1 } } \right) $$
$$\Rightarrow \left| ^{ n }{ C }_{ r } \right| =\left| ^{ n }{ C }_{ { r }^{ 1 } } \right| $$
So, $$n=r+{ r }^{ 1 }$$
$$n=\cfrac { 2n+10 }{ 5 } +\cfrac { 2n+5 }{ 5 } $$
$$5n=4n+15$$
$$n=15$$
The value of $${ }^nC_0 - { }^nC_1 + { }^n C_2 - .... + (-1)^{n^n}C_n$$ is:
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$$1$$
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$$0$$
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$$2^n$$
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$$n$$
Explanation
By Binomial Expansion we know,
$${ \left( a+b \right) }^{ n }=^{ n }{ C }_{ 0 }{ a }^{ 0 }{ b }^{ n }+^{ n }{ C }_{ 1 }{ a }^{ 1 }{ b }^{ n-1 }+....+^{ n }{ C }_{ n-1 }{ a }^{ n-1 }{ b }^{ 1 }+^{ n }{ C }_{ n }{ a }^{ n }{ b }^{ 0 }$$
Now putting $$a=-1$$ & $$b=+1$$
we have,
$${ \left( -1+1 \right) }^{ n }=^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 0 }{ \left( +1 \right) }^{ n-1 }+^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 1 }{ \left( +1 \right) }^{ n-1 }+.....+^{ n }{ C }_{ n-1 }{ \left( -1 \right) }^{ n-1 }\left( +1 \right) +^{ n }{ C }_{ n }{ \left( -1 \right) }^{ n }{ \left( +1 \right) }^{ 0 }$$
$$\Rightarrow 0 =^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 0 }{ \left( +1 \right) }^{ n-1 }+^{ n }{ C }_{ 0 }{ \left( -1 \right) }^{ 1 }{ \left( +1 \right) }^{ n-1 }+.....+^{ n }{ C }_{ n-1 }{ \left( -1 \right) }^{ n-1 }\left( +1 \right) +^{ n }{ C }_{ n }{ \left( -1 \right) }^{ n }$$
The value of the expression $${ _{ }^{ k-1 }{ C } }_{ k-1 }+{ _{ }^{ k }{ C } }_{ k-1 }+....{ _{ }^{ n+k-2 }{ C } }_{ k-1 }$$ is given by :
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$${ _{ }^{ n+k-1 }{ C } }_{ k-1 }$$
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$${ _{ }^{ n+k-1 }{ C } }_{ k }$$
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$${ _{ }^{ n+k }{ C } }_{ k }$$
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None of these
Explanation
We know,
$$\displaystyle \sum _{ m=k }^{ n }{ { { ^{ m }{ C }_{ r}=\ ^{ n+1 }{ C }_{ r+1\\ } } } } $$
According to the question,
$$ m= n+k-2$$
$$ r= k-1$$
Sum of series is given by
$$= \ ^{ m+1 }{ C }_{ k+1\\ }$$
$$ =\ ^{ n+k-1 }{ C }_{ k }$$
How many terms are there in the expansion of $${ \left( 1+2x+{ x }^{ 2 } \right) }^{ 10 }$$?
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$$11$$
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$$20$$
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$$21$$
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$$30$$
Explanation
Now,
$$(1+2x+x^{ 2 })^{ 10 }=((1+x)^{ 2 })^{ 10 }=(1+x)^{ 20 }$$
Now, the number of terms in the expansion of
$$(1+x)^{ n }$$ are $$n+1$$.
Thus, the number of terms in the expansion of $$(1+x)^{20}$$ will be $$20+1=21$$.
Hence, option C is correct.
Consider the expansion of $$(1+x)^{2n+1}$$
The sum of the coefficients of all the terms in the expansion is
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$$2^{2n-1}$$
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$$4^{n-1}$$
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$$2\times 4^n$$
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None of the above
Explanation
We can get the sum of all coefficients by putting $$x=1$$ in the expansion, because for calculating the coefficients we need the terms independent of $$x$$.
To find the sum of coefficients of all terms, put $$x=1$$ in the given expression$$(1+x)^{2n+1}$$ we get
$$2^{2n+1}=2.2^{2n}=2.4^n$$
Hence, C is the correct option.
What is n equal to ?
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$$5$$
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$$10$$
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$$15$$
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None of the above
Explanation
Using binomial theorem, we can write
$$\displaystyle \left({ x }^{ 3 }-\cfrac { 1 }{x^{2}} \right)^{ n }=\sum_{ r=0 }^{ n }\ { _{ r }^{ n }{ C{ \left(x ^{ 3 }\right) }^{ n-r } } } { \left(\cfrac { -1 }{ { x }^{ 2 } } \right) }^{ r }\\ =\displaystyle \sum_{ r=0 }^{ n }{ { \left(-1\right) }^{ r }\ {}_{ r }^{ n }{ C }{ x }^{ \left(3n-5r\right) } } $$
For $${x}^{3}$$ and $${x}^{10}$$ let value of r be $${ r }_{ 5 }$$ and $${ r }_{ 10 }$$ respectively
$$3n-5{r}_{10}=5$$ ......... $$(i)$$
$$3n-5{r}_{10}=10$$ .......... $$(ii)$$
$${ \left(-1\right) }^{ { r }_{ 5 } }\quad _{ { r }_{ 5 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 5 }\right) }$$$$=-{ \left(-1\right) }^{ { r }_{ 10} }\quad _{ { r }_{ 10 } }^{ n }{ C\quad }{ x }^{ \left(3n-5{ r }_{ 10}\right) }$$
Given sum of coefficient of $${x}^{5}$$ and $${x}^{10}$$ is 0
All terms are positive, but still there is a negative sign, means one of $${r}_{5}$$ and $${r}_{10}$$ is even and other is odd
By properties of combinations
$$_{ { r }_{ 5 } }^{ n }{ C}$$$$={}_{ { r }_{ 10 } }^{ n }{ C\quad }$$
$${r}_{5}+{r}_{10}=n$$ .......... $$(iii)$$
Adding $$(i)$$ and $$(ii)$$
$$6n-5\left({r}_{5}+{r}_{10}\right)=15$$
From $$(iii),$$
$${r}_{5}+{r}_{10}=n$$
$$\Rightarrow n=15$$
In the expansion of $$\left (x + \dfrac {1}{x}\right )^{n}$$, then the coefficient of the term indepenent of x is
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$$\dfrac {n!}{(r!)^{2}}$$
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$$\dfrac {n!}{(r + 1)! (r - 1)!}$$
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$$\dfrac {n!}{\left (\dfrac {n + r}{2}\right )! \left (\dfrac {n - r}{2}\right )!}$$
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$$\dfrac {n!}{\left [\left (\dfrac {n}{2}\right )!\right ]^{2}}$$
Explanation
In the expansion of $$\left (x + \dfrac {1}{x}\right )^{n}$$, $$r^{th}$$ term is
$$t_{r+1}={}^{n}C_{r} x^{n-r} \left(\dfrac{1}{x}\right)^{r}$$
$$\implies t_{r+1}={}^{n}C_{r} x^{n-r} x^{-r}$$
$$\implies t_{r+1}={}^{n}C_{r} x^{n-2r}$$ ... $$(i)$$
We need to find the coefficient of $$x^{0}$$
Then substitute $$n-2r=0\implies r=\dfrac{n}{2}$$
Substitute $$r$$ in RHS of $$(i)$$, we get
$$t_{r+1}={}^{n}C_{\frac{n}{2}}x$$
Then, coefficient of $$x$$ is $$\displaystyle ={}^{n}C_{\frac{n}{2}}=\dfrac{n!}{\left(\dfrac{n}{2}\right)! \left(n-\dfrac{n}{2}\right)!}=\dfrac{n!}{\left(\dfrac{n}{2}\right)!\left(\dfrac{n}{2}\right)!}=\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$$
Hence, coefficient of $$x^{0}=x$$ is
$$\dfrac{n!}{\left[\left(\dfrac{n}{2}\right)!\right]^{2}}$$
The sum of the series $$^{20}C_0 - \,^{20}C_1+\,^{20}C_2-\,^{20}C_3+...+\,^{20}C_{10}$$ is
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$$-^{20}C_{10}$$
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$$\dfrac{1}{2}\,^{20}C_{10}$$
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$$0$$
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$$^{20}C_{10}$$
Explanation
On putting $$x=-1$$ in
$$(1+x)^{20}=^{20}C_0 + ^{20}C_1x+... +{20}C_{10}x^{10} + ... + \,^{20} C_{20} x^{20}$$
We get,
$$0=\,^{20}C_0 - \,^{20}C_1+... - \,^{20}C_9+\,^{20}C_{10} - \, ^{20}C_{11}+ ... + \,^{20}C_{20}$$
$$\Rightarrow 0 = \,^{20}C_0 - \,^{20}C_1+ ... - \,^{20}C_9 + \,^{20}C_{10}-\,^{20}C_9+...+\,^{20}C_0$$ ........ $$[\because {}^{n}C_{r}={}^{n}C_{n-r}]$$
$$\Rightarrow 0 = 2(^{20}C_0-\,^{20}C_1+ ... -\,^{20}C_9)+\,^{20}C_{10}$$
$$\Rightarrow \,^{20}C_{10} = 2(^{20}C_0 - \,^{20}C_1+ ... + \,^{20}C_{10})$$
.... [Adding $${}^{20}C_{10}$$ on both the sides]
$$\Rightarrow \,^{20}C_0 - \,^{20}C_1+...+\,^{20}C_{10}=\dfrac{1}{2}\,^{20}C_{10}$$
If $$T_r=^{2016}C_rx^{2016-r}$$, for $$r=0, 1, ,....2016$$, then $$(T_0 - T_2+T_4....+T_{2016})^2+(T_1-T_3+T_5....T_{2015})^2$$ is equal to -
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$$(X^2-1)^{1008}$$
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$$(X+1)^{2016}$$
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$$(X^2-1)^{2016}$$
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$$(X^2+1)^{2016}$$
Explanation
Let $$i$$ represent iota.
$$(T_0-T_2+T_4+\dots+T_{2016}) = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})\dots(1)$$
Now, multiply $$(T_1-T_3+T_5+\dots-T_{2015})$$ with $$-i$$ to get $$-(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})\dots(2)$$
$$\implies (T_0-T_2+T_4+\dots+T_{2016})^2+(T_1-T_3+T_5+\dots-T_{2015})^2$$
$$= (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 + i^4(T_1-T_3+T_5+\dots-T_{2015})^2$$
$$ = (T_0+i^2T_2+i^4T_4+\dots+i^{2016}T_{2016})^2 -(iT_1+i^3T_3+i^3T_5+\dots+i^{2015}T_{2015})^2$$
$$ = (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$$ (As $$a^2-b^2 = (a+b)(a-b)$$)
Now, consider these terms seperately
$$(T_0+iT_1+i^2T_2+\dots)=(^{2016}C_0x^{2016}i^0 +^{2016}C_1x^{2015}i^1\dots) = (i+x)^{2016}$$
$$(T_0-iT_1+i^2T_2-i^3T_3\dots) = (^{2016}C_0(ix)^{2016} +^{2016}C_1(ix)^{2015}\dots) = (1+ix)^{2016}$$
$$\therefore (T_0+iT_1+i^2T_2+\dots+i^{2016}T_{2016}) (T_0-iT_1+i^2T_2-i^3T_3\dots)$$
$$= (i+x)^{2016}\times(1+ix)^{2016}$$
$$= [(i+x)(1+ix)]^{2016}$$
$$= [i-x+x+ix^2]^{2016}$$
$$= (1+x^2)^{2016}$$
If $$C_{0}, C_{1}, C_{2}, ...., C_{n}$$ are binomial coefficients of order $$n$$, then the value of $$\dfrac {C_{1}}{2} + \dfrac {C_{3}}{4} + \dfrac {C_{5}}{6} + .... =$$
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$$\dfrac {2^{n} + 1}{n + 1}$$
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$$\dfrac {2^{n} - 1}{n + 1}$$
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$$\dfrac {2^{n} + 1}{n - 1}$$
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$$\dfrac {2^{n}}{n + 1}$$
Explanation
$$(1+x)^{ n }=C_{ 0 }+C_{ 1 }x+C_{ 2 }x^{ 2 }+C_{ 3 }x^{ 3 }+......C_{ n }x^{ n }$$
Integrating, We have
$$ \dfrac { (1+x)^{ n+1 } }{ n+1 } =C_{ 0 }x+C_{ 1 }\dfrac { x^{ 2 } }{ 2 } +C_{ 2 }\dfrac { x^{ 3 } }{ 3 } +.....C_{ n }\dfrac { x^{ n+1 } }{ n+1 } $$
Putting, $$x=1$$
$$ \dfrac { (2)^{ n+1 } }{ n+1 } =C_{ 0 }+C_{ 1 }\dfrac { 1 }{ 2 } +C_{ 2 }\dfrac { 1 }{ 3 } +....$$
Putting, $$x=-1$$
$$ \dfrac { (0)^{ n+1 } }{ n+1 } =-C_{ 0 }+C_{ 1 }\dfrac { 1 }{ 2 } +C_{ 2 }\dfrac { (-1) }{ 3 } +....$$
Adding above two equations we have,
$$ \dfrac { (2)^{ n+1 } }{ n+1 } =2\left(\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....\right)$$
$$\implies \dfrac { (2)^{ n } }{ n+1 } =\dfrac { C_{ 1 } }{ 2 } +\dfrac { { C }_{ 3 } }{ 4 } .....$$
Hence, option D is correct.
The total number of terms in the expansion of $$(x + a)^{47} - (x - a)^{47}$$ after simplification is
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24
0%
47
0%
48
0%
96
Explanation
$${ \left( x+a \right) }^{ 47 }-{ \left( x-a \right) }^{ 47 }$$
When we expand the above equation using binomial expansion
$$ (x +y)^{n} = \displaystyle (\sum_{k=0}^{n} {^{n}C_k} x^{k}y^{n-k}) $$
So the above equation becomes
$$ (x +a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}a^{47-k}) $$
$$ (x -a)^{47} = \displaystyle (\sum_{k=0}^{47} {^{47}C_k} x^{k}(-a)^{47-k}) $$
$${ \left( x+a \right) }^{ 47 }\Rightarrow $$There are 48 terms in the expansion and all are positive
$${ \left( x-a \right) }^{ 47 }\Rightarrow $$There are 48 terms in the expansion
The terms with odd powers of a will be cancelled and those with even powers of a will add up.
24 terms will be positive and 24 negative in the expansion of $$ (x-a)^{47} $$
48 terms positive-[24 terms negative and 24 terms positive]
$$=48\quad terms\quad positive +24\quad terms\quad negative+24\quad terms\quad positive$$
$$=24$$ terms
Let $$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$$. Then
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$$a_{0} + a_{2} + ..... + a_{18} = a_{1} + a_{3} + ..... + a_{17}$$
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$$a_{0} + a_{2} + ..... + a_{18}$$ is even
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$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$9$$
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$$a_{0} + a_{2} + ..... + a_{18}$$ is divisible by $$3$$ but not by $$9$$
Explanation
Given :
$$((1 + x) + x^{2})^{9} = a_{0} + a_{1}x + a_{2}x^{2} + ..... + a_{18}x^{18}$$ ..... $$(i)$$
Put $$x=1$$ in $$(i)$$, we get
$$(1+1+1)^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$$
$${3}^{9}=a_{0}+a_{1}+a_{2}+.....+a_{18}$$ .... $$(ii)$$
Put $$x=-1$$ in $$(i)$$, we get
$$(1-1+1)^{9}=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$$
$$1=a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-..... -a_{17}+a_{18}$$ ..... $$(iii)$$
Adding $$(ii)$$ and $$(iii)$$, we get
$$3^{9}+1=a_{0}+a_{1}+....+a_{18}+a_{0}-a_{1}+a_{2}-.....-a_{17}+a_{18}$$
$$=2a_{0}+2a_{2}+2a_{4}+....+2a_{18}$$
$$\implies 2(a_{0}+a_{2}+a_{4}+....+a_{18})=3^{9} + 1$$
$$\implies a_{0} + a_{2} + ..... + a_{18} = \dfrac {3^{9} + 1}{2} \rightarrow even$$
Hence,
$$a_{0}+a_{2}+a_{4}+....+a_{18}$$ is even.
Let $$n \ge 5$$ and $$b \neq 0$$. In the binomial expansion of $${ \left( a-b \right) }^{ n }$$, the sum of the 5th and 6th terms is zero then $${ a }/{ b }$$ equals
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$$\dfrac { 5 }{ n-4 } $$
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$$\dfrac { 1 }{ 5\left( n-4 \right) } $$
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$$\dfrac{n-5}{6}$$
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$$\dfrac{n-4}{5}$$
Explanation
Solution:
Given that:
$$n\ge5,b\neq0,$$ given expansion is $$(a-b)^n$$ and
$$t_5+t_6=0$$
To find: $$\cfrac ab=?$$
Solution:
$$t_5={}^nC_4a^4(-b)^{(n-4)}$$ and
$$t_6={}^nC_5a^5(-b)^{(n-5)}$$
$$\because t_5+t_6=0$$
$$\therefore {}^nC_4a^4(-b)^{(n-4)}$$
$$+{}^nC_5a^5(-b)^{(n-5)}=0$$
$$\Longrightarrow {}^nC_5a={}^nC_4b$$
$$\Longrightarrow \cfrac ab=\cfrac{{}^nC_4}{{}^5C_5}=\cfrac{\cfrac{n!}{4!(n-4)!}}{\cfrac{n!}{5!(n-5)!}}=\cfrac{5}{n-4}$$
Hence, A is the correct answer.
In the expansion of $${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$$, the $$5^{th}$$ term from the end is
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$$\cfrac { 16486 }{ { x }^{ 8 } } $$
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$$\cfrac { 17010 }{ { x }^{ 8 } } $$
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$$\cfrac { 13486 }{ { x }^{ 8 } } $$
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None of these
Explanation
There are $$11$$ terms in the expansion of $${ \left( 3x-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 10 }$$
Therefore, $$5^{th}$$ term from the end.
$$=(10-5+2)^{th}$$ term from beginning
$$={ T }_{ 7 }={ _{ }^{ 10 }{ C } }_{ 6 }{ \left( 3x \right) }^{ 10-6 }{ \left( -\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ 6 }$$
$$=\cfrac { 10! }{ 6!4! } { 3 }^{ 4 }{ x }^{ 4 }\cfrac { 1 }{ { x }^{ 12 } }$$
$$ =\cfrac { 17010 }{ { x }^{ 8 } } $$
The coefficient of $$x^{49}$$ in the product $$(x - 1) (x - 2) (x - 3) .... (x - 50)$$ is
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$$-2250$$
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$$-1275$$
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$$1275$$
0%
$$2250$$
0%
$$-49$$
Explanation
Coefficient of $$x^{49}$$ in product of $$(x-1)(x-2)(x-3)(x-4)........(x-50)$$
We know that,
$$(x-1)(x-2)(x-3)(x-4)........(x-n)=x^n-(1+2+3+....+n)x^{n-1}+.......$$
Coefficient of $$x^{49}=-(1+2+3+...+50)$$
$$=-\cfrac{50\times51}{2}$$
$$=-1275$$
Hence, B is the correct option.
If $$C_0, C_1, C_2, C_3, .... $$ are binomial coefficients in the expansion of $$ ( 1+x)^n $$ , then $$ \dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac{}{} - \dfrac{}{} + ... $$ is equal to :
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$$ \dfrac{1}{n+1} - \dfrac{2}{n+2} + \dfrac{1}{n+3} $$
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$$ \dfrac{1}{n+1} + \dfrac{2}{n+2} - \dfrac{3}{n+3} $$
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$$ \dfrac{1}{n+2} - \dfrac{1}{n+1} + \dfrac{1}{n+3} $$
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$$ \dfrac{2}{n+1} - \dfrac{1}{n+2} + \dfrac{2}{n+3} $$
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$$ \dfrac{1}{n+2} - \dfrac{2}{n+1} + \dfrac{3}{n+3} $$
Explanation
We know
$$ (1-x)^n = C_0 - C_1 x + C_2x^2 - ... + (-1)^n C_n \cdot x^n $$
On multiplying both sides bby $$x^2$$ , we get
$$ (1-x)^n x^2 = C_0x^2 - C_1 x^3 + C_2x^4 - ... + $$
On integrating both sides by taking limit $$0$$ to $$1$$.
Therefore,
$$\displaystyle \int_0^1 ( 1-x)^n x^2 dx =\displaystyle \int_0^1 ( C_0x^2 - C_1x^3 + C_2x^4 + ....) dx$$
$$\displaystyle \int_0^1 x^n (1-x)^2 dx = \left[ C_0 \dfrac{x^3}{3} - C_1 \dfrac {x^4}{4} + C_2 \dfrac{x^5}{5} - ... \right]_0^1 $$
$$ \Rightarrow\displaystyle \int_0^1 x^n ( 1 + x^2 - 2x) dx = \dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ... $$
Here $$\dfrac {C_0}{3} - \dfrac {C_1}{4} + \dfrac {C_2}{5} - ... $$
$$ = \left[ \dfrac{x^{n+1}}{n+1} + \dfrac{x^{n+3}}{n+3} - \dfrac{2x^{n+2}}{n+2} \right]_0^1 $$
$$ = \left[ \dfrac {1}{n+1} + \dfrac{1}{n+3} - \dfrac {2}{n+2} \right] $$
The value of $$r$$ for which the coefficients of $$(r-5)$$th and $$(3r+1)$$th terms in the expansion of $${(1+x)}^{1/2}$$ are equal, is
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$$4$$
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$$9$$
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$$12$$
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None of these
Explanation
Since, coefficient of $$(r-5$$th term$$=$$ coefficient of $$=(3r+1)$$th term
$$\quad { _{ }^{ 12 }{ C } }_{ r-6 }=\quad { _{ }^{ 12 }{ C } }_{ 3r }$$
$$\Rightarrow$$ $$r-6=3r$$
or $$12-r+6=3r$$
$$\rightarrow$$ $$2r=-6$$ or $$4r=18$$
$$\Rightarrow$$ $$r=-3$$ or $$r=\cfrac { 18 }{ 4 } $$
Hence no value of $$r$$ exist, because $$r$$ neither be negative nor in fraction
If $$(1 + x + x^{2})^{n} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$$, then $$2a_{1} - 3a_{2} + ... -(2n + 1)a_{2n}$$ is equal to
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$$n$$
0%
$$-n$$
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$$n + 1$$
0%
$$-n - 1$$
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$$-n + 1$$
Explanation
Given,
$$(1 + x + x^{2})^{n} = 1 + a_{1}x + a_{2}x^{2} + ... + a_{2n}x^{2n}$$
$$\Rightarrow x(1 + x + x^{2})^{n} = x + a_{1}x^{2} + a_{2}x^{3} + ... + a_{2n}x^{2n + 1}$$
On diffferentiating w.r.t. $$x$$, we get
$$(1 + x +x^{2})^{n} + x\cdot n(1 + x + x^{2})^{n - 1}(1 + 2x)$$$$= 1 + 2a_{1}x + 3a_{2}x^{2} + ... + a_{2n} \cdot (2n + 1)x^{2n}$$
On putting $$x = -1$$, we get
$$(1 - 1 + 1)^{n} - n(1 - 1 + 1)^{n - 1}(1 - 2)$$$$= 1 - 2a_{1} + 3a_{2} + ... + a_{2n} (2n + 1)$$
$$\Rightarrow 1 - n(-1) = 1 - 2a_{1} + 3a_{2} + ... + a_{2n}(2n + 1)$$
$$\Rightarrow 2a_{1} - 3a_{2} .... =(2n + 1)a_{2n} = -n$$
The middle term in the expansion of $$\left( \dfrac{10}{x} + \dfrac{x}{10} \right )^{10}$$ is
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$$^{10}C_5$$
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$$^{10}C_6$$
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$$^{10}C_5 \dfrac{1}{x^{10}}$$
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$$^{10}C_5 x^{10}$$
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$$^{10}C_5 10^{10}$$
Explanation
MIddle term of the expansion of $$\left(\cfrac{10}x+\cfrac x{10}\right)^{10}$$ is $$T_6.$$
$$T_6={}^{10}C_{5}\left(\cfrac{10}x\right)^5\left(\cfrac x{10}\right)^5$$
$$T_6={}^{10}C_{5}$$
Hence, A is the correct option.
If the term free from $$x$$ in the expansion of $$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )^{10}$$ is $$405$$, then the value of $$k$$ is
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$$\pm 1$$
0%
$$\pm 3$$
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$$\pm 4$$
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$$\pm 2$$
Explanation
General term in the expansion of $$\left (\sqrt {x} - \dfrac {k}{x^{2}}\right )$$
$$T_{r + 1} = ^{10}C_{r} (\sqrt {x})^{10 - r} \left (\tfrac {-k}{x^{2}}\right )^{r}$$
$$= ^{10}C_{r}x^{\dfrac {10 - r}{2}} \cdot (-k)^{r} \cdot x^{-2r}$$
$$= ^{10}C_{r} (-k)^{r} x^{\left (\dfrac {10 - 5r}{2}\right )}$$
The term is free from $$x$$.
Put $$\dfrac {10 - 5r}{2} = 0$$
$$\Rightarrow r = 2$$
Now, $$^{10}C_{2} (-k)^{2} = 405$$
$$\Rightarrow \dfrac {10\times 9}{1\times 2}\cdot k^{2} = 405$$
$$\Rightarrow k^{2} = \dfrac {405}{45} = 9$$
$$\Rightarrow k = \pm 3$$.
Sum of coefficients of the last $$6$$ terms in the expansion of $${ \left( 1+x \right) }^{ 11 }$$ when the expansion is in ascending powers of $$x$$, is
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$$2048$$
0%
$$32$$
0%
$$512$$
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$$64$$
0%
$$1024$$
Explanation
$$(1+x)^{\prime \prime}=\sum_{r=0}^{11} n_{C_{r}}(1)^{n-r}(x)^{r} \\$$
$$\text { last } 6 \text { terms of this expansion are } \\$$
$$\qquad{ }^{\prime \prime} C_{11}+{ }^{\prime \prime} C_{10}+{ }^{\prime \prime} C_{q}+{ }^{\prime \prime} C_{8}+{ }^{\prime \prime} C_{7}+{ }^{\prime \prime} C_{6} \\$$
$$\text { using }{ }^{n} C_{r}+{ }^{n} C_{r+1}={ }^{n+1} C_{r+1}$$
$$=12 C_{11}+{ }^{12} C_{9}+{ }^{12} C_{7} \\$$
$$=12+220+792 \\$$
$$=1024$$
If $${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },\dots ,{ C }_{ 15 }$$ are binomial coefficients in $${ \left( 1+x \right) }^{ 15 }$$, then $$\dfrac { { C }_{ 1 } }{ { C }_{ 0 } } +2\dfrac { { C }_{ 2 } }{ { C }_{ 1 } } +3\dfrac { { C }_{ 3 } }{ { C }_{ 2 } } +\cdots +15\dfrac { { C }_{ 15 } }{ { C }_{ 14 } } $$ is equal to
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$$60$$
0%
$$120$$
0%
$$64$$
0%
$$124$$
0%
$$144$$
Explanation
We know that,
$$\dfrac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =\dfrac { n-\left( r-1 \right) }{ r } $$
$$\Rightarrow r\cdot \dfrac { ^{ n }{ { C }_{ r } } }{ ^{ n }{ { C }_{ r-1 } } } =n+1-r$$
$$\Rightarrow \displaystyle\sum _{ r=1 }^{ n }{ r\cdot \dfrac { ^{ 16 }{ { C }_{ r } } }{ ^{ 16 }{ { C }_{ r-1 } } } } =\displaystyle\sum _{ r=1 }^{ 16 }{ \left( 16-r \right) } $$
$$\displaystyle 16\times 15-\sum_{r=1}^{n}r$$
$$=16\times 15-\dfrac { 15\times 16 }{ 2 } $$
$$=240-120$$
$$=120$$
The middle term in the expansion of $${ \left( 1+x \right) }^{ 2n }$$ is
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$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n } }{ n! } $$
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$$\cfrac { 1.2.3....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$
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$$\cfrac { 1.3.5....(2n-1){ x }^{ n } }{ n! } $$
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$$\cfrac { 1.3.5....(2n-1){ 2 }^{ n }{ x }^{ n } }{ n! } $$
Explanation
Given expansion is $${ \left( 1+x \right) }^{ 2n }$$,
The total number of terms are $$2n+1$$ which is an odd number,
$$\therefore$$ there is only one middle term which is $$\left( \begin{matrix} 2n \\ n \end{matrix} \right) $$,
$$\left( \begin{matrix} 2n \\ n \end{matrix} \right) =\dfrac { \left( 2n \right) ! }{ n!n! } \\ $$
$$\Longrightarrow \dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n }n! \right] }{ n!n! } { x }^{ n }=\dfrac { \left[ 1.3.5...\left( 2n-1 \right) \right] \left[ { 2 }^{ n } \right] }{ n! } { x }^{ n }$$
If $$n\epsilon N$$ and $$(1 + 4x + 4x^{2})^{n} = \displaystyle \sum_{r = 0}^{r = 2n} a_{r}x^{r}$$ then value of $$2\displaystyle \sum_{r = 0}^{n}a_{2r}$$ equals
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$$9^{n} - 1$$
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$$9^{n} + 1$$
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$$3^{n }+ 1$$
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$$3^{n} - 1$$
Explanation
$$\because 1 + 4x + 4x^{2} = (1 + 2x)^{2}$$
$$\therefore (1 + 2x)^{2n} = \displaystyle \sum_{r = 0}^{2n}a_{r}x^{r}$$
putting $$x = 1$$
$$3^{2n} = \displaystyle \sum_{r = 0}^{2n} a_{r} = a_{0} + a_{1} + a_{2} + a_{3} + ..... + a_{2n} .... (i)$$
and putting $$x = -1$$
$$1 = \displaystyle \sum_{r = 0}^{r = 2n}a_{r} (-1)^{r} = a_{0} - a_{1} + a_{2} - a_{3} + .... + a_{2n} ... (ii)$$
Now adding (i) and (ii) we get
$$2(a_{0} + a_{2} + a_{4} + .... + a_{2n}) = 3^{2n} + 1$$
$$\Rightarrow 2\displaystyle \sum_{r = 0}^{r =n} a_{2r} = 9^{n} + 1$$
Hence choice (b) is correct answer.
The sum of the co-efficients of all odd degree terms in the expansion of $$\left(x + \sqrt {x^{3} - 1}\right)^{5} + (x - \sqrt {x^{3} - 1})^{5}, (x > 1)$$ is
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$$1$$
0%
$$2$$
0%
$$-1$$
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$$0$$
Explanation
$$(x+\sqrt {x^3-1})^5+(x-\sqrt {x^3-1})^5$$
$$(a+b)^5=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5 -------(i)$$
$$(a-b)^5=a^5-5a^4b+10a^3b^2-10a^2b^3+5ab^4-b^5 -------(ii)$$
$$(a+b)^5+(a-b)^5=2[a^5+10a^3b^2+5ab^4]$$
$$=2[x^5+10x^3(x^3-1)+5x(x^3-1)^4]$$
$$=2[x^5+10x^6-10x^3+5x(x^6-2x^3+1)]$$
$$=2x^5+20x^6-20x^3+10x^7-20x^4+10x$$
Here all the co-efficient of the above equation are odd terms
coefficients of even terms are cancelled out.
So, sum of coefficients $$\Rightarrow2+20-20+10-20+10=2$$
If $${C}_{r}$$ denotes the binomial coefficient $${ _{ }^{ n }{ C } }_{ r }$$ then $$\left( -1 \right) { C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+5{ C }_{ 2 }^{ 2 }+......(3n-1){ C }_{ n }^{ 2 }=\quad $$
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$$\left( 3n-2 \right) { _{ }^{ 2n }{ C } }_{ n }$$
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$$\left( \cfrac { 3n-2 }{ 2 } \right) { _{ }^{ 2n }{ C } }_{ n }$$
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$$\left( 5+3n \right) { _{ }^{ 2n }{ C } }_{ n }$$
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$$\left( \cfrac { 3n-5 }{ 2 } \right) { _{ }^{ 2n }{ C } }_{ n+1 }$$
Explanation
The general term of above series is
$$T_{r} = (3r-1)C_{r}^{2} $$
$$(-1)C_{0}^{2}+2C_{1}^{2}+5C_{2}^{2}+...+(3n-1)C_{n}^{2} $$
$$=\sum_{r=0}^{r=n}(3r-1)C_{r}^{2} $$
$$=3\sum_{r=0}^{r=n}rC_{r}^{2} - \sum_{r=0}^{r=n}C_{r}^{2}$$
$$=3S_{1} - S_{2}$$
The binomial expansion of $$(1+x)^n$$ and $$(1+\dfrac{1}{x})^n$$ is given by
$$(1+x)^n = C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}$$ ....[1]
$$(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}} = \sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$$
....[2]
To find the value of $$S_{2}$$, we can multiply above two equations and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [2], we get
$$(1+x)^n (1+\dfrac{1}{x})^{n} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}}$$
$$\implies \dfrac{(1+x)^{2n}}{x^{n}} = \sum_{r=0}^{r=n}C_{r}x^{r}.\sum_{r=0}^{r=n}C_{r}\dfrac{1}{x^{r}} $$
$$\implies \dfrac{(1+x)^{2n}}{x^{n}}=C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}$$ + terms containing x
Therefore, comparing coefficients of $$x^{0}$$ in L.H.S and R.H.S, we get
Coefficients of $$x^{0}$$ in R.H.S = C
oefficients of $$x^{0}$$ in L.H.S
$$\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}=$$ coefficient of $$x^{0}$$ in
$$\dfrac{(1+x)^{2n}}{x^{n}} $$
$$\implies C_{0}^{2}+C_{1}^{2}+C_{2}^{2}+...+C_{n}^{2}= ^{2n}C_{n} = S_{2}$$ ....[3]
$$(1+\dfrac{1}{x})^n = C_{0}+C_{1}\dfrac{1}{x}+C_{2}\dfrac{1}{x^{2}}+....+C_{n}\dfrac{1}{x^{n}}$$
Differentiating with respect to x, we get
$$\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{-1}{{x^{2}}}) = 0+C_{1}\dfrac{-1}{x^{2}}+C_{2}\dfrac{-2}{x^{3}}+....+C_{n}\dfrac{-n}{x^{n+1}}$$
$$\implies n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}}) = C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}}$$ ....[4]
To find the value of $$S_{1}$$, we can multiply equations [1] and [4], and then compare the coefficients of terms independent of x.
Multiplying equation [1] and [4], we get
$$n(1+\dfrac{1}{x})^{n-1}(\dfrac{1}{{x}})(1+x)^n=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )$$
$$\implies n\dfrac{(x+1)^{2n-1}}{x^{n}}=(C_{1}\dfrac{1}{x}+C_{2}\dfrac{2}{x^{2}}+....+C_{n}\dfrac{n}{x^{n}})( C_{0}+C_{1}x+C_{2}x^{2}+....+C_{n}x^{n} )$$
Therefore, comparing coefficients of $$x^{0}$$ in L.H.S and R.H.S, we get
Coefficients of $$x^{0}$$ in R.H.S = C
oefficients of $$x^{0}$$ in L.H.S
$$\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}=$$ coefficient of $$x^{0}$$ in $$n\dfrac{(x+1)^{2n-1}}{x^{n}} $$
$$\implies C_{1}^{2}+2C_{2}^{2}+3C_{3}^{2}+...+nC_{n}^{2}= n(^{2n-1}C_{n}) = S_{1} $$
Therefore,
$$3S_{1} - S_{2} = 3n(^{2n-1}C_{n})-^{2n}C_{n}$$
$$= 3n\dfrac{(^{2n}C_{n})}{2}-^{2n}C_{n}$$
$$= (\dfrac{3n-2}{2})^{2n}C_{n}$$
Hence, the answer is option (B)
Given $$(1-2x+5x^2-10x^3)(1+x)^n=1+a_1x+a_2x^2+...$$ and that $$a_1^2=2a_2$$ then the value of $$n$$ is-
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0%
6
0%
2
0%
5
0%
3
Explanation
Given that $$\left( 1-2x+5{ x }^{ 2 }-10{ x }^{ 3 } \right) { \left( 1+x \right) }^{ n }=1+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...$$
Given that $${ { a }_{ 1 } }^{ 2 }=2{ a }_{ 2 }$$
Here $${ a }_{ 1 }$$ is the coefficient of $$x$$ and $${ a }_{ 2 }$$ is the coefficient of $${ x }^{ 2 }$$,
$$\Longrightarrow { a }_{ 1 }=n-2$$ and $$\Longrightarrow { a }_{ 2 }=-2n+{ C }_{ 2 }+5$$
Substituting these values in the given relation we get,
$$\Longrightarrow { \left( n-2 \right) }^{ 2 }=2\left( 5-2n+\frac { n\left( n-1 \right) }{ 2 } \right) \\ \Longrightarrow n=6\\ $$
If $${ C }_{ 0 },{ C }_{ 1 },{ C }_{ 2 },.....{ C }_{ r }$$ are binomial coefficients in the expansion of $${(1+x)}^{n}$$ then
$${ C }_{ 1 }-\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } -\cfrac { { C }_{ 4 } }{ 4 } +....{ \left( -1 \right) }^{ n-1 }\cfrac { { C }_{ n } }{ n } $$ equals
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$$\sum _{ r=1 }^{ n }{ { \left( -1 \right) }^{ n-1 } } \cfrac { { C }_{ r } }{ r } $$
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$$\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } } $$
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$$\sum _{ r=2 }^{ n }{ \cfrac { 1 }{ r-1 } } $$
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None of these
Explanation
Consider $${ (1-x) }^{ n }={ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-{ C }_{ 3 }{ x }^{ 3 }+......$$
$$\Rightarrow 1-{ (1-x) }^{ n }={ C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-......$$
integrating both side w.r.t $$x$$ with limit $$0$$ to $$1$$
$$\displaystyle \therefore \int _{ 0 }^{ 1 }{ \left( { C }_{ 1 }x-{ C }_{ 2 }{ x }^{ 2 }+{ C }_{ 3 }{ x }^{ 3 }-..... \right) dx } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ (1-x) }^{ n } }{ 1-(1-x) } } dx$$
$$\displaystyle \Rightarrow \cfrac { { C }_{ 1 } }{ 1 } -\cfrac { { C }_{ 2 } }{ 2 } +\cfrac { { C }_{ 3 } }{ 3 } +....{ (-1) }^{ n }-1\cfrac { { C }_{ n } }{ n } =\int _{ 0 }^{ 1 }{ \cfrac { 1-{ x }^{ n } }{ 1-x } } dx\quad $$
$$\displaystyle \left( Using\quad \int _{ 0 }^{ a }{ f(x) } dx=\int _{ 0 }^{ a }{ f(a- } x)dx \right) $$
$$\displaystyle =\int _{ 0 }^{ 1 }{ \left( 1+x+{ x }^{ 2 }+...+{ x }^{ n-1 } \right) dx } $$
$$=1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } +\cfrac { 1 }{ 4 } +...\cfrac { 1 }{ n } =\sum _{ r=1 }^{ n }{ \cfrac { 1 }{ r } } $$
Prove that the coefficient of middle term in the expansion of $$ ( 1 + x ) ^{2n} $$ is equal to the sum of the coefficient of two middle terms in $$(1 + x)^{2n - 1}$$.
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0%
True
0%
False
Explanation
Statement is true, so option $$A$$ is correct.
Proof :- Middle term of $$(1 + X)^{2n}$$
Since $$2n$$ is even middle term = $$\left(\dfrac{2n}{2} + 1\right)^{th} = (n + 1)^{th}$$ term = $$T_{n + 1}$$
we know that general & term of expansion $$(a + b)^n$$ is -
$$T_{n + 1} = \,^nC_r \, a^{n - r} b^r$$
here, $$r = n , n = 2n , a = 1 , b = x$$
putting values
$$T_{nH} = \, ^{2n}C_n (1)^{2n - n} (x)^n$$
$$T_{nH} = \,^{2n}C_n (x)^n......(A)$$
Here, coefficient is $$^{2n} C_n$$
$$\Rightarrow$$ middle term of $$(1 + x)^{2n - 1}$$
Since, $$(2n - 1) $$ is add
There will be 2 middle terms
$$\dfrac{(2n - 1) + 1}{2} $$ and $$\left(\dfrac{(2n - 1) + 1}{2} + 1 \right) $$ term
$$= n^{th}$$ term and $$(n + 1)^{th}$$ term
$$= T_n $$ and $$T_{n + 1}$$
we know that general term for expansion $$(a + b)^n$$
$$T_{rH} = \,^nC_r \, a^{n - r} \, b^r$$
for $$T_n$$ in $$(1 + x)^{2n - 1}$$
Putting $$r = n - 1, n = 2n - 1, a = 1 \, \& \, b = x$$
$$T_n = \, ^{2n - 1} C_{n - 1} \, (1)^{2n - 1 - ( n - 1)} x_{( n- 1)} = \, ^{2n - 1}C_{n - 1} x^{(n - 1)}$$
Coefficient pf middle term $$(n^{th} \, term) = \, ^{2n - 1}C_{n - 1}$$
$$T_{nH} = \, ^{2n - 1}C_n \, (1)^{2n - 1 - n} x^{(n)} = \, ^{2n - 1} C_{n} \, (X)^n$$
Sum of the coefficent of the middle term in the expansion of $$(1 + X)^{2n - 1}$$
$$= \,^{2n -1}C_{n - 1} + \, ^{2n - 1}C_n$$
$$= \dfrac{(2n- 1)!}{(2n - 1 - n + 1)! (n -1)!} + \dfrac{(2n - 1)!}{(2n - 1 - n)! n!}$$
$$= \dfrac{(2n - 1)!}{n! (n -1)!} + \dfrac{(2n - 1)!}{(n - 1)! n!}$$
$$= \dfrac{2 (2n - 1)!}{n! (n - 1)!} \times \dfrac{n}{n}$$
$$= \dfrac{2n (2n - 1)!}{n! n (n - 1)!} = \dfrac{(2n)!}{n! \, n!} = \, ^{2n}C_n .....(B)$$
Equation $$(A)$$ = Equation $$(B)$$
State true or false.
The middle term in the expansion of
$$\left(\dfrac{x}{a} \, - \, \dfrac{a}{x}\right)^{10}=-252$$
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0%
True
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False
Find the term of the expansion of $$\displaystyle\, \left ( \sqrt[3]{x^{-2}} + x \right )^7$$ containing $$x$$ in the second power.
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$$T_4$$
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$$T_5$$
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$$T_6$$
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$$T_7$$
Explanation
Note:- The general term or $$(r+1)^{th}$$ term of $$(a+b)^{n}$$ is $${T}_{r+1}=^{ n }{ C }_{ r }{a}^{n-r}{b}^{r}$$
$${T}_{r+1}=^{ 7 }{ C }_{ r }\left( \sqrt [ 3 ]{ { x }^{ -2 } } \right) ^{ 7-r }\left(x \right)^{r}$$
$$\Rightarrow {T}_{r+1}= ^{ 7 }{ C }_{ r }\left({x}^{-2}\right)^{{7-r}/{3}}{x}^{r}$$
$$\Rightarrow {T}_{r+1}=^{ 7 }{ C }_{ r } {x}^{{-2}/{3}(7-r)+r}$$
Now, if power of $$x$$ must be $$2$$, then $$r$$ is:-
$$\therefore {x}^{{-2}/{3}(7-r)+r}= {x}^{2}$$
$$\Rightarrow \cfrac{-2}{3}(7-r)+r=2$$
$$\Rightarrow \cfrac{-14}{3}+\cfrac{2}{3}r+r=2$$
$$\Rightarrow \cfrac{5r}{3}= \cfrac{20}{3}$$
$$\Rightarrow r=4$$
Therefore $${T}_{4+1}={T}_{5}$$ is the term which contains $${x}^{2}$$
Sum of coefficients in the expression of $$(x+2y+z)^{10}$$ is
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$$2^{10}$$
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$$3^{10}$$
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1
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None of these
Explanation
Sum of = coeff. $$4^{10}$$
The $$6^{th}$$ coefficient in the expansion of $$\left (2x^2 - \dfrac {1}{3x^2}\right)^{10}$$
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$$-\dfrac{986}{27}$$
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$$\dfrac{986}{27}$$
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$$\dfrac{896}{27}$$
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$$- \dfrac{896}{27}$$
Explanation
$${ \left( 2{ x }^{ 2 }-\dfrac { 1 }{ { { 3x }^{ 2 } } } \right) }^{ 10 }$$
$$6^{th}$$ coefficient $$\rightarrow$$ $$5^{th}$$ term
$$={ ^{ 10 }{ C } }_{ 5 }{ \left( 2{ x }^{ 2 } \right) }^{ 5 }{ \left( \dfrac { -1 }{ { { 3x }^{ 2 } } } \right) }^{ 5 }$$
$$={ ^{ 10 }{ C } }_{ 5 }\times 2^5\times \dfrac{-1}{3^5}$$
$$=-\dfrac{252\times 32}{243}$$
$$=-\dfrac{896}{27}.$$
Hence, the answer is $$-\dfrac{896}{27}.$$
Prove that $$C_0+C_1+C_2+.....C_n=2^n$$
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$$2^n$$
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$$2^{n-1}$$
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$$2^{n+1}$$
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None of the above.
Explanation
$$(1+x)^n=^nC_0+^nC_1x+\cdots+^nC_nx^n$$
Let $$x=1$$
$$\Rightarrow C_0+C_1.1+C_2.1^2+\cdots+C_n.1^n=(1+1)^n=2^n$$
$$\Rightarrow C_0+C_1+C_2.1+\cdots+C_n=(1+1)^n=2^n$$
In the expansion of $$(\sqrt[5]{3}+\sqrt[7]{2})^{24}$$, the rational term is
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$$T_{14}$$
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$$T_{16}$$
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$$T_{15}$$
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$$T_7$$
the coefficients of $$x^{49}$$ in the polynomial.
$$\left (x \, - \, \dfrac{C_1}{C_0}\right) \, \left (x \, - \, 2^2 \, \dfrac{C_2}{C_1}\right) \, \left (x \, - \, 3^2 \, \dfrac{C_3}{C_2}\right) \, ..... \, \, \left (x \, - \, 50^2 \, \dfrac{C_{50}}{C_{49}}\right)$$ is
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$$50\displaystyle \sum _{ r=1 }^{ 50 }{ r-{ r }^{ 2 } } $$
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$$51\displaystyle \sum _{ r=1 }^{ 50 }{ { r }^{ 2 } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r } $$
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$$51\displaystyle \sum _{ r=1 }^{ 50 }{ { r } } -\displaystyle \sum _{ r=1 }^{ 50 }{ r^{ 2 } } $$
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$$none\ of\ these$$
Explanation
We Know that $$(x \, - \, a) \, (x \, - \, b) \,(x \, - \, c).....50$$
factors $$ = \, x^{50}\, x^{49} \, . S_1 \, + x^{48} \, S_2 \, -....$$
The coefficient of $$x^{49}$$ is
$$S_1 \, = \,- \,\left [ \dfrac{C_1}{C_0} \, + \, 2^2 \, \dfrac{C_2}{C_1} \,+ \,3^2 \, \dfrac{C_3}{C_2}\right] \, + .....r^2 \,\dfrac{C_r}{C_{r \, - \, 1}}]$$
$$ Now \, r^2 \, .\dfrac{C_r}{C_{r \, - \, 1}} \, = \, r^2 \, .\dfrac{n \, -r \, + \, 1}{r}$$
$$= r \, (51 \, - \, r) \, = \, 51r \, - r^2$$
$$ \, \therefore \, \sum_{r \, = \, 1}^{50} \, r^2 \, \dfrac{c_r}{C_{r \, - \, 1}} \, 51 \, \sum_{r \, = \, 1}^{50} \, r \, - \, \sum_{r \, = \, 1}^{50} \, r^2$$
$$= \, 51. \, \dfrac{N (N \, + \, 1)}{2}\, - \, \dfrac{N(N \, + \, 1)\, (2N \, + \, 1)}{6}$$
$$= \, 51. \, \dfrac{50. \, 51)}{2}\, - \, \dfrac{50. \ 51. \, (101)}{6}$$
$$= \,\dfrac{1}{6} \, 50. \, 51 \, \left [ 3 \, \times \, 51 \, - \, 101 \right ] \, = \, 25 \, \times \, 17 \, \times \, 52$$
If the sum of the co-efficient in the expansion of $$(a+b)^n$$ is $$1024$$, then the greatest co-efficient in the expansion is
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$$252$$
0%
$$352$$
0%
$$452$$
0%
$$552$$
Explanation
Sum of coefficient in the expansion of $$(a+b)^{n}$$ is $$1024$$
put $$a=b=1,2^{n}=1024\implies n=10$$
The greatest coefficient is $$^{10}C_{5}=252$$
If $$n\ge 2$$ then
$$3.{ C }_{ 1 }-4.{ C }_{ 2 }+5.{ C }_{ 3 }-......+{ \left( -1 \right) }^{ n-1 }\left( n+2 \right) .{ C }_{ n }$$ is equal to
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$$-1$$
0%
$$2$$
0%
$$-2$$
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$$1$$
Explanation
$$(1+x)^n=^nC_0+^nC_1x+^nC_2x^2+....+^nC_n x^n$$
$$x^2(1+x)^n=^nC_0 x^2+^nC_1x^3+^nC_2x^4+....+^nC_n x^{n+2}$$
differentiate w.r.t $$x$$
$$2x(1+x)^n+nx^2(1+x)^{n-1}=2^nC_0 x+3 ^nC_1x^2+4 ^nC_2x^3+....+(n+2)^nC_n x^{n+1}$$
put $$x=-1$$
$$0=-2+3 ^nC_1-4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n$$
so $$3 ^nC_1- 4 ^nC_2+5 ^nC_3+.....+(-1)^{n-1}(n+2) ^nC_n=2$$
The sum $$^{10}C_3 + ^{11}C_3 + ^{12}C_3 + .... + ^{20}C_3$$ is equal to
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$$^{21}C_4$$
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$$^{21}C_4 - ^{10}C_4$$
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$$^{21}C_4 - ^{11}C_4$$
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$$^{21}C_{17}$$
Explanation
$$ ^{10}C_{3} +\ ^{11}C_{3} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} $$ ....(i)
addition and subtraction $$^{10}C_{4}$$ in equation (i)
$$ ^{10}C_{4} -\ ^{10}C_{4} +\ ^{10}C_{3} +\ ^{11}C_{3} + --- +\ ^{20}C_{3}$$
[we know that -
$$ ^nc_{r-1} +\ ^nc_{r} =\ ^{n+1}c_{r}] $$
$$^{10}C_{3} +\ ^{10}C_{4} +\ ^{11}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$=\ ^{11}C_{4} +\ ^{11}C_{3} +\ ^{12}C_{3} + ---+\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$=\ ^{12}C_{4} +\ ^{12}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4} $$
$$ =\ ^{13}C_{4} +\ ^{14}C_{3} + --- +\ ^{20}C_{3} -\ ^{10}C_{4}$$
Similarly
,
Continue
So, we get
$$^{21}C_{4} - \ ^{10}C_{4}$$
So, option (B) is correct
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Practice Class 11 Engineering Maths Quiz Questions and Answers
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