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CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 12 - MCQExams.com
CBSE
Class 11 Engineering Maths
Binomial Theorem
Quiz 12
If $$C_r \, = \, (^{100 C _ r }) , then \, E = \sum_{r = 0 }^{n + 4 } (-1)^r \, c_r \, c_{r + 1}$$
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$$(^{100} C _ {45} )$$
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$$(^{100} C _ {47} )$$
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$$(^{101} C _ {50} )$$
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$$(^{100 C _ {51} })$$
Explanation
$$(1 \, + \, x)^{101} \, = \, C_0 \, + \, C_1 x \, + \, C_2 \, x^2 \, + \, .....$$
$$(1 \, - \, x)^{101} \, = \, C_0 \, x^{101} \, - C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....$$
Multiplying both sides,
$$(x^2 \, - \, 1)^{101} \, = \, (C_0 \, + \, C_1 \, x \, + \, C_2 \, x^2 \, + \, ....)$$
$$(C_0 \, x^{101} \, - \, C_1 \, x^{100} \, + \, C_2 \, x^{99} \, - \, ....)$$ ....(1)
Now E = $$C_0 \, C_1 \, - \, C-1 \, C_2 \, + \, C_2 \, C_3 \, - \, ....$$
= -[Coefficient of $$x^{100}$$ in the product of R.H.S. of (1)]
or = -[Coefficient of $$(x^2)^{50}$$ in the product of R.H.S. of (1)] ....(2)
Now
$$(x^2 \, - \, 1)^{101} \, = \, (-1)^{101} \, (1 \, - \, x^2)^{101} \, = \, -(1 \, - \, x^2)^{101}$$
$$\therefore \, \, \, $$Coefficient of $$(x^2)^{50}$$
= - $$(- 1)^{50} \, \, ^{101}C_{50} \, = \, - ^{101}C_{50}$$ ...(3)
$$\therefore \, \, \, \, E \, = \, (-) \, (-) \, ^{101}C_{50} \, = \, ^{101}C_{50} \, by \, (2) \, and \, (3)$$
If the $$rth$$ term in the expansion of $$\left(\dfrac{x}{3}-\dfrac{2}{x^{2}}\right)^{10}$$ contains $$x^{4}$$ then $$r$$ is equal to
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$$2$$
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$$1$$
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$$3$$
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$$5$$
If $$ C_{0}$$,$$C_{1}$$,$$C_{2}$$
,
...,$$C_{n} $$ are the binomial coefficients and $$n$$ is odd, then
$$ 2C_{1} $$+$$2^{3}.C_{5}+$$...$$+2^{n}\ ^{n}C_{n}$$ equals
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$$\dfrac {3^{n}+(-1)^{n}}{2}$$
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$$\dfrac {3^{n}-(-1)^{n}}{2}$$
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$$\dfrac {3^{n}+1}{2}$$
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$$\dfrac {3^{n}-1}{2}$$
$$\begin{array} { l } { \text { Assertion } ( \mathrm { A } ) : \text { The expansion of } ( 1 + x ) ^ { n } = } \\ { C _ { 0 } + C _ { 1 } x + C _ { 2 } x ^ { 2 } + \ldots + C _ { n } x ^ { n } } \\ { \text { Reason (R): If } x = - 1 , \text { then the above expansion is } } \\ { \text { zero } } \end{array}$$
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$$\begin{array} { l } { \text { Both A and R are true and } \mathrm { R } \text { is the correct } } \\ { \text { explanation of } \mathrm { A } } \end{array}$$
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$$\begin{array} { l } { \text { Both A and R are true and R is not the } } \\ { \text { correct explanation of } \mathrm { A } } \end{array}$$
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$$A$$ is true, but $$R$$ is false
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$$A$$ is false, but $$R$$ is true
Explanation
We Know that Assertion is nothing but the standard binomial expansion of
$$(1+x)^n$$$$=C_{0}+C_{1}X+C_{2}X^2+ \dots +C_{n}x^n$$
So if $$x=-1$$ then
$$(1+(-1))^n=(1-1)^n=0$$
$$Ans:$$ $$Opt:[B]$$
If the ratio $$T_2 : T_3$$ in the expansion of $$(a+b)^n$$ and $$T_3 : T_4$$ in the expansion of $$(a+b)^{n+3}$$ are equal , then n $$=$$
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3
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4
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5
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6
The coefficient of $${x^5}$$ in the expansion of $${\left( {1 + {x^2}} \right)^5}{\left( {1 - x} \right)^4}$$ is
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$${4.^6}{C_4}$$
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$${2.^6}{C_4}$$
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$${2.^6}{C_2}$$
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$${4.^6}{C_2}$$
In the binomial expansion of $$( a - b ) ^ { n } n > 0$$ and the sum $$5 ^ { t h }$$ and $$6 ^ { \text { th } }$$ terms is zero, then $$\frac { a } { b }$$ equal to
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$$\frac { 5 } { n - 4 }$$
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$$\frac { 6 } { n - 5 }$$
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$$\frac { n - 5 } { 6 }$$
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$$\frac { n - 4 } { 5 }$$
The co-efficient of $$x$$ in the expansion of $$\left( 1 - 2 x ^ { 3 } + 3 x ^ { 5 } \right) \left( 1 + \dfrac { 1 } { x } \right) ^ { 8 }$$ is
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$$56$$
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$$65$$
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$$154$$
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$$62$$
Explanation
$$(1-2x^3+3x^5) \left (1+\dfrac {1}{x}\right)^8$$ find coefficient of $$x$$
$$\Rightarrow \left( 1-2{ x }^{ 3 }+3{ x }^{ 5 } \right) \left( 1+8\left( \dfrac { 1 }{ x } \right) +_{ }^{ 8 }{ { C }_{ 2 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 2 }+_{ }^{ 8 }{ { C }_{ 3 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 3 }+_{ }^{ 8 }{ { C }_{ 4 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 4 }+_{ }^{ 8 }{ { C }_{ 5 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 5 }+_{ }^{ 8 }{ { C }_{ 6 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 6 }+_{ }^{ 8 }{ { C }_{ 7 }^{ } }{ \left( \dfrac { 1 }{ x } \right) }^{ 7 }+{ \left( \dfrac { 1 }{ x } \right) }^{ 8 } \right) $$ (By Binomial theorem)
$$=(1+(2.^BC_2 +3^8C_4)x+-----)$$
$$=$$ co-efficient of $$x$$ is
$$-2\times ^8C_2+3\times ^8C_4=\dfrac {-2\times 8!}{6!\ 2!}+3\times \dfrac {8!}{4!\ 4!}$$
$$=\dfrac {-2\times 8\times 7}{2\times 1}dfrac {3\times 8\times 7\times 6\times 5}{4\times 3\times 2\times 1}=-56+210$$
$$=154$$
If the sum of odd terms and the sum of even terms in $$(x+a)^{n}$$ are $$p$$ and $$q$$ respectively then $$p^{2}+q^{2}=$$
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$$\dfrac{(x+a)^{2n}-(x-a)^{2n}}{2}$$
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$$(x+a)^{2n}-(x-a)^{2n}$$
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$$\dfrac{(x+a)^{2n}+(x-a)^{2n}}{2}$$
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$$(x+a)^{2n}+(x-a)^{2n}$$
The sum of the series
$$\dfrac {2\left(\dfrac {n}{2}\right)!\left(\dfrac {n}{2}\right)!}{(n!)}[C^{2}_{0}-2C^{2}_{1}+3C^{2}_{2}...+(-1)^{n}(n+1)C^{2}_{n}]$$
where $$n$$ is an even positive integer, is equal to
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$$0$$
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$$(-1)^{\dfrac {n}{2}}(n+1)$$
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$$(-1)^{\dfrac {n}{2}}(n+2)$$
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$$(-1)^{n}n$$
Integral part of $$(8+3\sqrt{7})^{n}$$ is
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an even number
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an odd number
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an even or odd number depending upon the value of n
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nothing can be said
If $$A$$ and $$b$$ are coefficients of $$( 1 + x ) ^ { 2 }$$ and $$( 1 + x ) ^ { 2 n - 1 }$$ respectively, then
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$$d = B$$
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$$A = 2 B$$
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$$2 A = B$$
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$$A + B = 0$$
The total number of terms in the expansion of $$( x + a ) ^ { 200 } + ( x - a ) ^ { 200 }$$ after simplification is
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$$101$$
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$$102$$
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$$201$$
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$$202$$
The sum of the co-efficients of all odd degree terms in the expansion of $${ \left( x+\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }+{ \left( x-\sqrt { { x }^{ 3 }-1 } \right) }^{ 5 }$$
$$(x> 1)$$ is:
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$$2$$
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$$-1$$
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$$0$$
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$$1$$
Explanation
Sum of the coefficient of odd term is given by
$$= 2\left[_{ }^{ 5 }{ { C }_{ 0 }^{ } } + {_{ }^{ 5}}{ { C }_{ 2 }^{ } } + {_{ }^{ 5 }}{ { C }_{ 4 }^{ } }\right]$$
$$=2\left[1+10-10+5-10+5\right]$$
$$=2(1+5+5-10)$$
$$=2$$.
Sum of the coefficient of integral powers of x in $${\left(1-2\sqrt{x}\right)}^{50}$$ is
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$$\dfrac{{3}^{50}+1}{2}$$
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$$\dfrac{{3}^{50}}{2}$$
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$${2}^{49}-1$$
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$${2}^{49}+1$$
Find the middle terms(s) in the expansion of $$\left ( 3x-\dfrac{2}{x^{2}} \right )^{15}$$.
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$$\dfrac {-6435\times 3^7\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}$$
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$$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^8}{x^9}$$
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$$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^7 \times 2^7}{x^9}$$
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$$\dfrac {-6435\times 3^8\times 2^7}{x^6}, \dfrac {6437\times 3^8 \times 2^8}{x^9}$$
Sum of coefficients of $${ x }^{ 2r }$$, $$r= 1,2,3,$$....... in $$(1+x{ ) }^{ n }$$ is
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$$({ 2 }^{ n-1 }-1)$$
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$$({ 2 }^{ n-1 }+1)$$
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$$({ 2 }^{ n-2 }+1)$$
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$$({ 2 }^{ n-2 }-1)$$
The sum of coefficients of integral powers of $$x$$ in the binomial expansion of $$(1-2\sqrt x)^{50}$$ is
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$$\frac{1}{2}(3^{50}-1)$$
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$$\frac{1}{2}(2^{50}+1)$$
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$$\frac{1}{2}(3^{50}+1)$$
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$$\frac{1}{2}(3^{50})$$
Coefficient of $$x^{n}$$ in expansion of $$\dfrac{(1+2x)^{2}}{(1-x)^{3}}$$ is
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$$2n$$
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$$\dfrac{3}{2}(3n^{2}+n)$$
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$$n^{2}+n-1$$
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$$None\ of\ these$$
Coefficient of $$x^{r}$$ in the expansion of $$(1-2x)^{-1/2}$$ is
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$$\dfrac{(2r)!}{(r!)^{2}}$$
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$$\dfrac{(2r)!}{2^{r}(r!)^{2}}$$
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$$\dfrac{(2r)!}{(r!){2^{2r}}}$$
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$$\dfrac{(2r)!}{2^{r}(r+1)!(r+1)}$$
Explanation
Coefficient of $${x^r}$$ in the expansion of
$${\left( {1 - 2x} \right)^{ - \frac{1}{2}}}$$
by using binomial expansion
$$\begin{array}{l} \Rightarrow \frac { { \left[ { \frac { 1 }{ 2 } \left( { \frac { 1 }{ 2 } +1 } \right) \left( { \frac { 1 }{ 2 } +2 } \right) ...........\left( { \frac { 1 }{ 2 } +r-1 } \right) \left( { 2x } \right) } \right] } }{ { \left( { r! } \right) } } \\ \Rightarrow \frac { { \frac { 1 }{ 2 } .\frac { 3 }{ 2 } .\frac { 5 }{ 2 } .........\left( { r-\frac { 1 }{ 2 } } \right) { 2^{ r } }{ x^{ r } } } }{ { \left( { r! } \right) } } =\frac { { 2r! } }{ { { { \left( { r! } \right) }^{ 2 } } } } \end{array}$$
The coefficient of $$x^{99}$$ in $$(x+1)(x+3)(x+5).....(x+199)$$ is
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$$1+2+3+...+99$$
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$$1+3+5+...+199$$
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$$1.3.5............199$$
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$$None\ of\ these$$
Explanation
$$\text x^{99} \text { in }(x+1)(x+3)(x+5) \ldots(x+199) \\$$
$$\text { Multiplying }(x+1)(x+3) \\$$
$$\text { will give } x^{2}+(1+3) x+(1)(3) \\$$
$$\text { so coefficient } x \text { is }(1+3) \\$$
$$\Rightarrow \operatorname{2n}(x+1)(x+3)(x+5)$$
So coefficient of $$x^{2}=(1+3+5)$$
So coefficient of $$x^{n}$$ in $$(n+1)$$ terms will be sum of coefficients
So we can say it will be sum of roots of equation
So coefficient of $$x^{99}$$ will be roots in 100 terms
$$1+3+5+\ldots+199$$
The sum of coefficient of integral powers of $$x$$ in the binomial expansions $${\left( {1 - 2\sqrt x } \right)^{50}}$$ is:
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$$\frac{1}{2}\left( {{3^{50}} - 1} \right)$$
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$$\frac{1}{2}\left( {{2^{50}} + 1} \right)$$
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$$\frac{1}{2}\left( {{3^{50}} + 1} \right)$$
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$$\frac{1}{2}\left( {{3^{50}}} \right)$$
The middle term in the expansion of $${\left(3x-\dfrac{{x}^{3}}{6}\right)}^{9}$$ is
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$$\dfrac{21}{16}{x}^{19}$$
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$$\dfrac{-21}{16}{x}^{19}$$
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$$\dfrac{21}{16}{x}^{-19}$$
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$$\dfrac{-21}{16}{x}^{-19}$$
The sum of the co-efficient of all odd degree terms in the expansion of $$\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right)$$
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$$0$$
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$$1$$
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$$2$$
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$$- 1$$
Sum of the series $$)^{100}C_1)^2 + 2(^{100}C_2)^2 + 3(^{100}C_3)^2+.......+100(^{100}C_{100})^2$$ equals
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$$\frac{2^{99}[1.3.5 ... ... (199)]}{99!}$$
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$$100. ^{200}C_{100}$$
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$$50. ^{200}C_{100}$$
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$$100. ^{199}C_{99}$$
The sum of the series $$^{2020}C_0- ^{2020}C_1+^{2020}C_2-^{2020}C_3+.....+^{2020}C_{1010}$$ is
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$$\dfrac{1}{2}^{2020}C_{1010}$$
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$$^{2020}C_{1010}$$
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Zero
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$$\dfrac{-1}{2}^{2020}C_{1010}$$
Explanation
$$(1-x)^{m}={ }^{n} C_{0}-{ }^{n} C_{1} x+\ldots+{ }^{n} C_{n} x^{n}\\$$
$$\text { Put } x=1 \text { and } n=2020\\$$
$$0={ }^{2020} C_{0}-{ }^{2020} C_{1}+{ }^{2020} C_{2}+\ldots { }^{2020} C_{2020}\\$$
$$\text { also }^{n} C_{r}={ }^{n} C_{n-r}\\$$
$$\text { }^{2020} C_{2020}={ }^{2020} \mathrm{Co}\\$$
$$\left[2\left[{ }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots\right]+{ }^{2020} \mathrm{C}_{1010}\right]=0\\$$
$$\text { So } { }^{2020} \mathrm{C}_{0}-{ }^{2020} \mathrm{C}_{1}+\ldots-{ }^{2020} \mathrm{C}_{1009}=\\$$
$$\frac{-1}{2}^{2020} C_{1010}$$
The value of $$\dfrac{1}{12!}+\dfrac{1}{10!2!}+\dfrac{1}{8!4!}+...+\dfrac{1}{12!}$$
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$$\dfrac{{2}^{12}}{12!}$$
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$$\dfrac{{2}^{11}}{12!}$$
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$$\dfrac{{2}^{11}}{11!}$$
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None of these
If $${ \left( 1+x+2{ x }^{ 2 } \right) }^{ 20 }=0+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+......+{ a }_{ 40 }{ x }^{ 40 }$$ then $${ a }_{ 1 }+{ a }_{ 3 }+{ a }_{ 5\quad }+......+{ a }_{ 37\quad }$$ equals -
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$${ 2 }^{ 19}({ 2 }^{ 20 }-21)$$
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$${ 2 }^{ 20 }({ 2 }^{ 19 }-19)$$
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$${ 2 }^{ 19 }({ 2 }^{ 20 }+21)$$
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None of these
The greatest terms of the expansion $$(2x+5y)^{13}$$ when $$x=10$$, $$y=2$$ is?
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$$^{13}C_5\cdot 20^8\cdot 10^5$$
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$$^{13}C_6\cdot 20^7\cdot 10^4$$
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$$^{13}C_4\cdot 20^9\cdot 10^4$$
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None of these
The coefficient of x$$^9$$ in (x - 1) (x - 4) (x - 9)........(x - 100) is
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-235
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235
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385
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None of these
Find the value of $$\dfrac{1}{\left(n-1\right)!}+\dfrac{1}{\left(n-3\right)!3!}+\dfrac{1}{\left(n-5\right)!5!}+...$$
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$$\dfrac{{2}^{n-1}}{(n-1)!}$$
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$$\dfrac{{2}^{n}}{n!}$$
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$$\dfrac{{2}^{n-1}}{n!}$$
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None of these
The constant term in the expansion of $${ (1+x) }^{ n }{ (1+\frac { 1 }{ x } ) }^{ n }$$ is
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$${ C }_{ 0 }^{ 2 }+2{ C }_{ 1 }^{ 2 }+3{ C }_{ 2 }^{ 2 }+......+(n+1){ C }_{ n }^{ 2 }$$
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$${ ({ C }_{ 0 }+{ C }_{ 1 }+.....+{ C }_{ n }) }^{ 2 }$$
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$${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+.....+{ C }_{ n }^{ 2 }$$
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None of these
The largest coefficient in the expansion of $$\left(4+3x\right)^{25}$$ is
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$$^{25}C_{11}3^{25}\left(\dfrac{4}{3}\right)^{14}$$
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$$^{25}C_{11}4^{25}\left(\dfrac{3}{4}\right)^{11}$$
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$$^{25}C_{14}4^{14}3^{11}$$
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$$^{25}C_{14}4^{11}.3^{14}$$
Explanation
$$=(4+3 x)^{25}\\$$
$$\left({}^{25} c_{0}(4)+{ }^{25} c_{1}(4)^{2}(3 x)+\cdots^{25} c_{25}\right.\\$$
$$(4+3 x)^{25}\\$$
$$\text { Let Tr the term is largest }\\$$
$$\therefore \quad T_{r-1}<T_{r}>T_{r+1}\\$$
$$=\frac{T_{2}}{T_{r_{-1}}}>1 \quad \frac{T_{r+1}}{T_{r}}<1$$
$$T_{r+1^{}}= {}^{25} C_{r}(4)^{25-r}(3 x)^{k}$$
$$T_{r-1}=25_{C_{r-2}}(4)^{25-r+2}(3 x)^{r-2}$$
$$T_{r}={ }^{25} C_{r-1}\left(4\right)^{26-r}(3 x)^{r-1}$$
$$\therefore \frac{25 C_{r-1}(4)^{26-r}(3 x)^{r-1}}{25 C_{r-2}(4)^{27-r}(3 x)^{r-2}}>1$$
$$\frac{(25-r)}{(x-1)} \times \frac{3}{4}>1$$
$${}75-3 r > 4 r -4\\ $$
$$79>7 r\\$$
$$11.1>r$$
$$\Rightarrow r=12$$
largest term = $$\left({ }^{25}c_{11}(3)^{25}\left(\frac{4}{3}\right)^{14}\right.)$$
The sum of the co-efficient of all odd degree terms in the expansion of $$\left( x + \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } + \left( x - \sqrt { x ^ { 3 } - 1 } \right) ^ { 5 } , ( x > 1 )$$ is :
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$$- 1$$
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$$1$$
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$$0$$
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$$2$$
If the last term in the binomial expansion of $$\left(2^{1/3}-\dfrac {1}{\sqrt {2}}\right)^{n}$$ is $$\left(\dfrac {1}{3^{5/3}}\right)^{\log_{3}8}$$, then the $$5^{th}$$ terms form the beginning is:
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$$210$$
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$$420$$
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$$103$$
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$$None\ of\ these$$
Coefficient of $$x^{25}$$ in $$(1+x+x^{2}+x^{3}+....+x^{10})^{7}$$ is
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$$^{31}C_{15}-7.^{20}C_{14}$$
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$$^{31}C_{14}-7.^{20}C_{14}$$
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$$31$$
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$$None\ of\ these$$
The coefficient of $$x^{8}$$ in $$(1+2x^{2}-x^{3})^{9}$$ is
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$$1680$$
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$$2140$$
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$$2520$$
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$$2730$$
The coefficients of $$x^{10}$$ in the expansion of $$(1+x)^{15}+(1+x)^{16}+(1+x)^{17}+....+(1+x)^{30}$$ is
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$$^{31}C_{10}-^{15}C_{10}$$
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$$^{31}C_{11}-^{15}C_{11}$$
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$$^{30}C_{10}-^{15}C_{10}$$
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$$^{31}C_{10}-^{14}C_{11}$$
The sum of the coefficient in the expansion of $$(a+2b+c)^{11}$$ is-
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$$4^{11}$$
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$$32$$
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$$31$$
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$$None\ of\ these$$
State true or false.
The general term for $$3,7,13,21,31,43 $$........ is $$n^2−(n−1),n=1,2,3,...$$
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True
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False
The coefficient of $$x^{3}$$ in the expansion of $$(1+2x+3x^{2})^{10}$$ is
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Less than $$200$$
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Less than $$400$$ but greater than $$200$$
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$$1400$$
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$$1500$$
The coefficient of $${x}^{n}$$ in the expansion of $$\dfrac { 1 }{ \left( 1-x \right) \left( 1-2x \right) \left( 1-3x \right) }$$ is
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$$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+3 }+1 \right)$$
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$$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 2 }^{ n+3 }+1 \right)$$
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$$\dfrac { 1 }{ 2 } \left( { 2 }^{ n+2 }-{ 3 }^{ n+2 }+1 \right)$$
0%
$$None\ of\ these$$
Explanation
$$\text { We know that } \\$$
$$\frac{1}{1-a} \text { is sum of } \infty GP=1+a+a^{2}+a^{3} \ldots\infty \\$$
$$\text { Similary: }$$
$$\frac{1}{(1-x)(1-2 x)(1-3 x)}={\left(1+x+x^{2}+\ldots\infty\right)\left(1+(2 x)+(2 x)^{2}\ldots\infty\right)}\\ $$
$${\left(1+(3 x)+(3 x)^{2}+(3 x)^{3}\ldots\infty\right.)}$$
From this equation, we observe that
coefficient of $$x^{n}$$ is $$: \rightarrow$$
$$\frac{1}{2}\left[2^{n+2}-3^{n+3}+1\right]$$
Option (a)
The number of rational terms in the expansion of $${ \left( 1+\sqrt { 2 } +\sqrt [ 3 ]{ 3 } \right) }^{ 6 }$$ is
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$$6$$
0%
$$7$$
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$$5$$
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$$8$$
The coefficient of $$x^4$$ in the expansion of $$(1+x+x^2+x^3)^{11}$$ is
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$$990$$
0%
$$495$$
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$$330$$
0%
none of these
Explanation
$$=\left(1+x+x^{2}+x^{3}\right)^{11}$$
$$=\left((1+x)\left(1+x^{2}\right)\right)^{11}$$
$$=(1+x)^{ 11}\left[1+^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots^{11} c_{11}\left(x^{2}\right)^{11}\right.]$$
$$=\left[1+{ }^{11} c_{1} x+{ }^{11} c_{2} x^{2}+\cdots\right]\left[1+{ }^{11} c_{1} x^{2}+{ }^{11} c_{2} x^{4}+\cdots\right]$$
$$={ }^{11} c_{0} \cdot{ }^{11} c_{2} \cdot x^{4}+{ }^{11} c_{2} \cdot{ }^{11} c_{1} \cdot x^{4}+{ }^{11} c_{4}{ }^{11} c_{0} x^{4}$$
$$=\quad 990 x^{4}$$
Option $$(A)$$
The coefficient of x$$^{24}$$ in the expansion of
(1 +3x + 6x$$^2$$ + 10x$$^3$$+ -----------+$$\infty$$)$$^{2/3}$$ =
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300
0%
250
0%
25
0%
205
Explanation
$$\left(1+3 x+6 x^{2}+10 x^{3}+\ldots\infty\right)^{\frac{2}{3}} \\$$
$$\text { We know that, } \\$$
$${ (1+\alpha) }^{n}=1+n \alpha+\frac{n(n-1) \alpha^{2}}{2 !}+\ldots \\$$
$$\text { comparing with given eqn, }$$
$$\frac{n(n-1)}{2} \cdot \alpha^{2}=6 x^{2} \\$$
$$=n(n-1) \frac{9 x^{2}}{n^{2}}=6 x^{2} \\$$
$$\Rightarrow n=-3$$
$$n \alpha=3 x$$
$$\Rightarrow\alpha=\frac{3 x}{n}$$
$$\alpha=-x$$
$$\therefore \text {Expansion is} (1-x)^{-\not 3 \times \frac{2}{\not3}}$$
$$\quad =(1-x)^{-2}$$
$$=1+2 x+\frac{2 \times 3}{2 !} x^{2}+\frac{2 \times 3 \times 4}{3 !} x^{3}+\ldots$$
$$\therefore$$ coeff. of $$x^{24}$$ is 2
$$= \frac{2 \times 3 \times 4 \times\ldots25}{24 !}$$
$$=25$$
The co-efficient of $$x^{k}$$ in expansion of $$1+\left(1+x\right)+\left(1+x\right)^{2}++\left(1+x\right)^{n}$$ is : $$\left(n>k\right)$$
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$$^{n}C_{k}$$
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$$^{n+1}C_{k}$$
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$$^{n+1}C_{k+1}$$
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$$None\ of\ these$$
The number of terms in the expansion of $${ \left[ { a }^{ 3 }+\dfrac { 1 }{ { a }^{ 3 } } +1 \right] }^{ 100 }$$ is
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$$201$$
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$$300$$
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$$200$$
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$$^{ 100 }{ C }_{ 3 }$$
Explanation
$$\left(a^{3}+\frac{1}{a^{3}}+1\right)^{100} \\$$
$$\frac{1}{a^{300}}\left(1+a^{3}+a^{6}\right)^{100}$$
Here, terms will be in the form,
$$a^{0} ; a^{3}, a^{6}\cdots$$
and highest power term will be $$\left(a^{6}\right)^{100}$$
This forms a A.P of powers,
$$\therefore \quad 600=0+(n-1) 3$$
$$\therefore \quad n=201 \\$$
$$\text { Option (a) }$$
For $$x\in R, x\neq -1$$ if $$(1+x)^{2016}+x(1+x)^{2015}+x(1+x)^{2014}+.+x^{2016}=\sum _{ i=0 }^{ 2016 }{ { a }_{ i }{ x }^{ i } }$$, then $$a_{17}$$ is equal to
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$$\dfrac{2017!}{ 17!2000!}$$
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$$\dfrac{2016!}{ 17!1999!}$$
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$$\dfrac{2017!}{2000!}$$
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$$\dfrac{2016!}{ 16!}$$
The middle term in the expansion of $$\left(1-\dfrac{1}{x}\right)^{n}\left(1-x\right)^{n}$$ is
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$$^{2n}C_{n}$$
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$$-^{2n}C_{n}$$
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$$-^{2n}C_{n-1}$$
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$$none\ of\ these$$
If the middle term in the expansion of $$( 1 + x ) ^ { 2 n }$$ is the greatest term, then $$x$$
lies in the interval ___________________.
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$$\left( \frac { n } { n + 1 } , \frac { n + 1 } { n } \right)$$
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$$\left( \frac { n + 1 } { n } , \frac { n } { n + 1 } \right)$$
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$$( n - 2 , n )$$
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$$( n - 1 , n )$$
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