If $$(1+x)^{2n} =a_0+a_1x....+a_{2n}x^{2n}$$, then

$$a_1+a_2+a_4.....=\dfrac 12 (a_0+a_1+a_2.....)$$

MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 11

The value of

$x$ in the expression

${ \left( x+{ x }^{ \log _{ 10 }{ x } } \right) }^{ 5 }$ , if the third term in the expansion is

$10,00,000,$ is

0%
${ 10 }^{ -1 }$
0%
${ 10 }^{ 1 }$
0%
${ 10 }^{ -5/2 }$
0%
${ 10 }^{ 5/2 }$

Explanation

Substitute

$\log _{ 10 }{ x=z }$

Then, given expression

$={ \left( x+{ x }^{ z } \right) }^{ 5 }$

Now,

${ T }_{ 3 }=^{ 5 }{ { C }_{ 2 } }{ x }^{ 3 }{ \left( { x }^{ z } \right) }^{ 2 }=10{ x }^{ 3+2z }={ 10 }^{ 6 }$

$\therefore \quad { x }^{ 3+2z }={ 10 }^{ 5 }$

Taking log. we get

$\left( 3+2z \right) \log _{ 10 }{ x } =5\log _{ 10 }{ 10 }$

$\Rightarrow \left( 3+2z \right) z=5\Rightarrow 2{ z }^{ 2 }+3z-5=0$

$\displaystyle \Rightarrow \left( z-1 \right) \left( 2z+5 \right) =0\Rightarrow z=1,-\frac { 5 }{ 2 }$

$\therefore \log _{ 10 }{ x } =1$ or

$\displaystyle -\frac { 5 }{ 2 }$

$\therefore x={ 10 }^{ 1 }$ or

$\displaystyle { 10 }^{ -\tfrac { 5 }{ 2 } }$

The total number of terms which are dependent on the value of

$x$ in the expansion of

$\left(x^2 - 2 + \displaystyle\frac{1}{x^2}\right)^n$ is equal to

0%
$2n + 1$
0%
$2n$
0%
$n$
0%
$n + 1$

Explanation

$(x^2-2+\dfrac{1}{x^2})^{n}$

$=((x-\dfrac{1}{x})^2)^{n}$

$=(x-\dfrac{1}{x})^{2n}$

$T_{r+1}=\:{10}C_{r}x^{2n-2r}$
For term independent of

$x$

$2n-2r=0$

$n=r$
Hence there will be

$1$ term independent of

$x$ .
Since the total number of terms are

$2n+1$ .
Hence the total number of term dependent on

$x$ will be
Total number of terms-(total number of terms independent of

$x$ ).

$=2n+1-1$

$=2n$

$c_{0},c_{1},c_{2}\cdots c_{n}$ are binomial coefficients in

$\left ( 1+x \right )^{n}$ , then the value of

$c_{1} + c_{5} + c_{9}+c_{13}+\cdots$ equals

0%
$\displaystyle 2^{n-1} + 2^{\frac{n}{2}}\sin \left ( \frac{n\pi }{4} \right )$
0%
$\displaystyle 2^{n-1} + 2^{\frac{n}{2}}\cos \left ( \frac{n\pi }{4} \right )$
0%
$\displaystyle \frac{1}{2}\left ( 2^{n-1}+2^\tfrac{n}{2}\:\sin \frac{n\pi }{4} \right )$
0%
$\displaystyle \frac{1}{2}\left ( 2^{n-1}-2^\tfrac{n}{2}\:\sin \frac{n\pi }{4} \right )$

Explanation

$c_{0},c_{1},c_{2}\cdots c_{n}$ are binomial coefficients of

$\left ( 1+x \right )^{n}$
i.e

$\left ( 1+x \right )^{n} = c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+\dots$
put

$x=i$

$\Rightarrow \left ( 1+i \right )^{n} = c_{0}+c_{1}i-c_{2}-c_{3}i+\dots$

$\Rightarrow \displaystyle 2^{\frac{n}{2}}c$ is

${\displaystyle \frac{n\pi}{4}} = c_{0}+c_{1}i-c_{2}-c_{3}i+\dots$
Comparing imaginary parts gives

$\displaystyle 2^{\frac{n}{2}}\sin {\dfrac{n\pi}{4}} = c_{1}-c_{3}+c_{5}+\dots$ ------(1)

$2^{n-1} = c_{1}+c_{3}+c_{5}+\dots$ ------(2) (

$\because\quad c_{0}+c_{1}+c_{2}+\dots = 2^{n}$ )

$\therefore$ (1)+(2) gives

$c_{1}+c_{5}+c_{9}+\dots = \displaystyle\frac{1}{2}\left(2^{n-1}+\displaystyle 2^{\frac{n}{2}}\sin {\dfrac{n\pi}{4}}\right)$

The expresion

$^{45}C_{8}$ +

$\sum _{ k=1 }^{ 7 }{^{ 52-k} C_{ 7 }}$ +

$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 50-i }}$

0%
$^{55}C_{7}$
0%
$^{57}C_{8}$
0%
$^{57}C_{7}$
0%
None of these

Explanation

$^{45}C_{8}$ +

$\sum _{ k=1 }^{ 7 }{^{ 52-k} C_{ 7 }}$ +

$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 50-i }}$

$\Rightarrow$

$^{45}C_{8}$ +

$^{45}C_{7}$ +

$^{46}C_{7}$ +......+

$^{51}C_{7}$ +

$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 57-i-(50-i) }}$

$\Rightarrow$ (

$^{45}C_{8}$ +

$^{45}C_{7}$ )+

$^{46}C_{7}$ +......+

$^{51}C_{7}$ +

$\sum _{ i=1 }^{ 5 }{^{ 57-i} C_{ 7) }}$

$\Rightarrow$ (

$^{46}C_{8}$ +

$^{46}C_{7}$ )+......+

$^{51}C_{7}$ +

$^{52}C_{7}$ +

$^{53}C_{7}$ +........+

$^{56}C_{7}$
:
:

$\Rightarrow$

$^{56}C_{8}$ +

$^{56}C_{7}$

$\Rightarrow$

$^{57}C_{8}$

The coefficient of

$x^{n-2}$ in the polynomial

$(x-1)(x-2)(x-3)....(x-n)$ is

0%
$\displaystyle \frac{n(n^{2}+2)(3n+1)}{24}$
0%
$\displaystyle \frac{n(n^{2}-1)(3n+2)}{24}$
0%
$\displaystyle \frac{n(n^{2}+1)(3n+4)}{24}$
0%
None of these

Explanation

There are total of $n$ brackets. The term $x^{n-2}$ will be formed when integers are chosen from any two brackets and $x$ is chosen from all the other brackets and multiplied.

Thus, the Coefficient of $x^{n-2}$ is

$C= (1\times2+1\times3+...+1\times n)+(2\times3+2\times4+..+2\times n)+...+((n-1)\times n)$

$= \left(\dfrac {(n)(n+1)}{2}-1\right)+2\left(\dfrac {(n)(n+1)}{2}-1-2\right)+...+(n-1)\left(\dfrac {(n)(n+1)}{2}-(1+2+3+..+(n-1))\right)$

$= \left\{(1+2+3+...+(n-1))(\dfrac {(n)(n+1)}{2})\right\} - \left\{ 1+2(1+2)+3(1+2+3)+...+(n-1)(1+...+n-1)\right\}$

$= \left\{(\dfrac {(n-1)(n)}{2})(\dfrac {(n)(n+1)}{2})\right\} - \left\{\sum_{1}^{n-1} k(\dfrac {(k)(k+1)}{2})\right\}$

$= \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \left\{ \sum_{1}^{n-1} \dfrac {k^{3}+k^{2}}{2} \right\}$

$= \left\{ \dfrac {n^{2}(n^{2}-1)}{4}\right\} - \dfrac {1}{2}\left\{ (\dfrac {(n-1)(n)}{2})^{2} + \dfrac {(n-1)n(2n-1)}{6} \right\}$

$= \dfrac {n(n-1)}{4} \left\{ {n(n+1)} - \dfrac {n(n-1)}{2} - \dfrac {2n-1}{3} \right\}$

$= \dfrac {n(n-1)}{4}\left\{ \dfrac {n(n+3)}{2} -\dfrac {2n-1}{3} \right\}$

$= \dfrac {n(n-1)}{4}\left\{ \dfrac {3n^{2}+9n-4n+2}{6} \right\}$

$= \dfrac {n(n-1)}{4}\left\{ \dfrac {(3n+2)(n+1)}{6} \right\}$

$\therefore$ Option $B$ is correct.

The value of the expression

$\displaystyle \frac{1 + 4\:.\:343 + 7\:.\:4 + 2\:.\:3\:.\:49 + 7\:.\:343}{16 + 2^6\:.\:3^1 + 2^{5}\:.\:3^{3} + 2^{6}\: . \: 3^{3} + 2^{4}\:.\:3^{4}}$ equal

0%
$\displaystyle 1$
0%
$\displaystyle 2$
0%
$\displaystyle 4$
0%
$\displaystyle 3$

Explanation

$\displaystyle \frac{1 + 4\:.\:343 + 7\:.\:4 + 2\:.\:3\:.\:49 + 7\:.\:343}{16 + 2^6\:.\:3^1 + 2^{5}\:.\:3^{3} + 2^{6}\: . \: 3^{3} + 2^{4}\:.\:3^{4}}$

Above expression can be written as

$\dfrac{\>^4C_0\cdot1^4 + \>^4C_1\cdot1^3\cdot7 + \>^4C_2\cdot 1^2\cdot 7^2 + \>^4C_3\cdot 1\cdot 7^3 + \>^4C_4\cdot7^4}{\>^4C_0\cdot2^4 + \>^4C_1\cdot2^3\cdot6 + \>^4C_2\cdot 2^2\cdot 6^2 + \>^4C_3\cdot 2\cdot 6^3 + \>^4C_4\cdot6^4}$

$= \dfrac{(1+7)^4}{(2+6)^4} = 1$

The value of

$\displaystyle B=\sum_{0\leq r\leq s\leq n}^{}$

$\displaystyle \sum \left ( C_{r}-C_{s} \right )^{2}$ is

0%
$\displaystyle \left ( n+1 \right )^{n}-2^{2n}$
0%
$\displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{n}$
0%
$\displaystyle \left ( n+1 \right )\displaystyle ^{2n}C_{n}-2^{2n}$
0%
$\displaystyle 2^{2n}-2^{n}$

Explanation

Given

$\displaystyle A=\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$ ....(1)

and

$\displaystyle \sum _{ r=0 }^{ n } C_{ r }^{ 2 }=^{ 2n }C_{ n }$ ....(2)

Given expansion can be written as

$\left( \sum _{ r=0 }^{ n } C_{ r } \right) ^{ 2 }=(C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }$

Consider,

$(C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }$

$\Rightarrow \displaystyle (C_{ 0 }+C_{ 1 }+C_{ 2 }+....+C_{ n })^{ 2 }=\sum _{ r=0 }^{ n } C_{ r }^{ 2 }+2\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$

We know that

$C_0+C_1+C_2+....+C_n = 2^n$

$\Rightarrow \displaystyle (2^n)^2=^{2n}C_{n}+2A$ (by (1), (2))

$\Rightarrow \displaystyle 2^{2n}=^{2n}C_n+2A$

$\Rightarrow \displaystyle 2A= 2^{2n}-^{2n}C_n$ .....(3)

Now, consider

$\displaystyle B=\sum_{0\leq r\leq s\leq n} \displaystyle \sum \left ( C_{r}-C_{s} \right )^{2}$

$\displaystyle ={ ({ C }_{ 0 }-{ C }_{ 1 }) }^{ 2 }+{ ({ C }_{ 0 }-{ C }_{ 2 }) }^{ 2 }+{ ({ C }_{ 0 }-{ C }_{ 3 }) }^{ 2 }+.....+{ ({ C }_{ 0 }-{ C }_{ n }) }^{ 2 }$

$+ { ({ C }_{ 1 }-{ C }_{ 2 }) }^{ 2 }+{ ({ C }_{ 1 }-{ C }_{ 3 }) }^{ 2 }+....+{ ({ C }_{ 1 }-{ C }_{ n }) }^{ 2 }$

$+{ ({ C }_{ 2 }-{ C }_{ 3 }) }^{ 2 }+{ ({ C }_{ 2 }-{ C }_{ 4 }) }^{ 2 }+...+{ ({ C }_{ 2 }-{ C }_{ n }) }^{ 2 }$

$+....+{ ({ C }_{ n-1 }-{ C }_{ n }) }^{ 2 }$

$={ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 1 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 1 } })+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 2 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 2 } })+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ 3 } })+..+{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 0 }{ { C }_{ n } })+$

${ { (C }_{ 1 } }^{ 2 }+{ { C }_{ 2 } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ 2 } })+{ { (C }_{ 1 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ 3 } })+....+{ { (C }_{ 1 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 1 }{ { C }_{ n } })+$

${ { (C }_{ 2 } }^{ 2 }+{ { C }_{ 3 } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ 3 } })+{ { (C }_{ 2 } }^{ 2 }+{ { C }_{ 4 } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ 4 } })+...+{ { (C }_{ 2 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ 2 }{ { C }_{ n } })+$

$....+{ { (C }_{ n-1 } }^{ 2 }+{ { C }_{ n } }^{ 2 }-2{ C }_{ n-1 }{ { C }_{ n } })$

$\Rightarrow \displaystyle B=n{ { (C }_{ 0 } }^{ 2 }+{ { C }_{ 1 } }^{ 2 }+...+{ { C }_{ n } }^{ 2 })-2\sum \sum _{ 0\leq r\leq s\leq n } C_{ r }C_{ s }$

$\Rightarrow B=n(^{2n}C_n)-2A$ (by(2))

$\Rightarrow B=n(^{2n}C_n)-2^{2n}+^{2n}C_n$ (by (3))

$\Rightarrow B=(n+1)^{2n}C_n -2^{2n}$

$C_{0},C_{1},C_{2}....,C_{n}$ denote the binomial coefficients in the expansion of

$\left ( 1+x \right )^{n}$ , then

$\cfrac{C1}{C0}+2\cfrac{C2}{C1}++3\cfrac{C3}{C2}+.....+n\cfrac{Cn}{Cn-1}$ equals

0%
$\displaystyle \frac{n}{2}$
0%
$\displaystyle \frac{n+1}{2}$
0%
$\displaystyle \frac{n\left ( n-1 \right )}{2}$
0%
$\displaystyle \frac{n\left ( n+1 \right )}{2}$

Explanation

Writing the general term, we get

$T_{r}=r.\dfrac{\:^{n}C_{r}}{\:^{n}C_{r-1}}$

$=r.\dfrac{\dfrac{n!}{(n-r)!.r!}}{\dfrac{n!}{(n-(r-1))!.(r-1)!}}$

$=r.\dfrac{n-(r-1))!.(r-1)!}{(n-r)!.r!}$

$=r.\dfrac{(n-(r-1)}{r}$

$=(n-(r-1))$

$=n+1-r$
Applying summation

$\sum _{r=1} r^{n} (n+1)-r$

$=n(n+1)-\dfrac{n(n+1)}{2}$

$=(n+1)(n-\dfrac{n}{2})$

$=\dfrac{n(n+1)}{2}$ .

If the fourth term of

${ \left( \sqrt { { x }^{ \left( \cfrac { 1 }{ 1+\log { x } } \right) } } +\sqrt [ 12 ]{ x } \right) }^{ 6 }$ is equal to 200 and

$x>1$ , then

$x$ is equal to

0%
$10\sqrt { 2 }$
0%
$10$
0%
${ 10 }^{ 4 }$
0%
$10/\sqrt { 2 }$

Explanation

${ \left( \sqrt { { x }^{ \left( \cfrac { 1 }{ 1+\log { x } } \right) } } +\sqrt [ 12 ]{ x } \right) }^{ 6 }$

$={ \left( { x }^{ \cfrac { 1 }{ 2(1+\log { x) } } }+{ x }^{ \frac { 1 }{ 12 } } \right) }^{ 6 }$

Now,

$T_{4}=T_{3+1}=^6C_{3}{ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } } {x}^{\cfrac{1}{4}}$

$\Rightarrow \displaystyle 200=20{ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 } }$

$\Rightarrow 10={ x }^{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 } }$
Taking log on both sides,

$\Rightarrow \displaystyle 1=\left[{ \cfrac { 3 }{ 2(1+\log { x) } } +\frac { 1 }{ 4 }}\right]log_{10} x$

$\Rightarrow \displaystyle \frac{1}{log x}-\frac{1}{4}=\frac{3}{2(1+\log x)}$

$\Rightarrow (\log x)^2+3\log x-4=0$

$\Rightarrow (\log x+4)(\log x -1)=0$

$\Rightarrow \log _{10}x=1$ (

$\because \log x=-4$ , not possible )

$\Rightarrow x=10$

The number of irrational terms in the expansion of

$(\sqrt[8]{5}+\sqrt[6]{2})^{100}$ is

0%
$97$
0%
$98$
0%
$96$
0%
$99$

Explanation

$T_{r+1}=\:^{100}C_{r}5^\left({\tfrac{100-r}{8}}\right)2^\left({\tfrac{r}{6}}\right)$
For rational terms,

$100 -r$ should be a multiple of

$8$ and

$r$ should be a multiple of

$6$

Hence we get rational terms for

$r=12, 36, 60$ and

$84$
Hence there will be

$4$ rational terms.
Therefore, total number of irrational terms will be

$101-4$

$=97$ terms.

The value of the expression

${ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-......+{{ \left( -1 \right) }^{ n }} \times {{ C }_{ n }^{ 2 }}$ is

0%
$0$ , if $n$ is odd
0%
${ \left( -1 \right) }^{ n }$ , if $n$ is odd
0%
${ \left( -1 \right) }^{ n/2 }\ { _{ }^{ n }{ C } }_{ n/2 }$ , if $n$ is even
0%
${ \left( -1 \right) }^{ n-1 }\ { _{ }^{ n }{ C } }_{ n-1 }$ , if $n$ is even

Explanation

Given ,

${ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-......{{ \left( -1 \right) }^{ n }} \times { C }_{ n }^{ 2 }\$
Case I : When

$n$ is odd, i.e.

$n=2m+1$ ,

$S={ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-.....+{{ \left( -1 \right) }^{ 2m }}\times { C }_{ 2m }^{ 2 }\quad +{{ \left( -1 \right) }^{ 2m+1 }} \times { C }_{ 2m+1 }$

$S={ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-.....+{ C }_{ 2m }^{ 2 } -{ C }_{ 2m+1 }^{ 2 }$

$=({ C }_{ 0 }^{ 2 }-{ C }_{ 2m+1 }^{ 2 })-({ C }_{ 1 }^{ 2 }-{ C }_{ 2m }^{ 2 })+....+{ \left( -1 \right) }^{ m }({ C }_{ m }^{ 2 }-{ C }_{ m+1 }^{ 2 })$
But

${ _{ }^{ 2m+1 }{ C } }_{ 0 }={ _{ }^{ 2m+1 }{ C } }_{ 2m+1 }\quad ;\quad { _{ }^{ 2m+1 }{ C } }_{ 1 }={ _{ }^{ 2m+1 }{ C } }_{ 2m }$ etc.
Therefore,

$S=0$
Case II: When

$n$ is even, i.e.

$n=2m$ .

$S={ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }-......+{{ \left( -1 \right) }^{ 2m }} \times { C }_{ 2m }^2$

$S={ C }_{ 0 }^{ 2 }-{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }......+{ C }_{ 2m }^2$
Consider the coefficient of the constant term in the following expression.

$\left[ { C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-.....+{{ \left( -1 \right) }^{ 2m }} \times { C }_{ 2m } { x }^{ 2m } \right] \left[ { C }_{ 0 }+{ C }_{ 1 }\cfrac { 1 }{ x } +{ C }_{ 2 }\cfrac { 1 }{ { x }^{ 2 } } +....+{ C }_{ 2m }\cfrac { 1 }{ { x }^{ 2m } } \right]$

$={ \left( 1-x \right) }^{ 2m }{ \left( 1+\cfrac { 1 }{ x } \right) }^{ 2m }$
Coefficient of

${x}^{2m}$ in

${ \left( 1-x \right) }^{ 2m }{ \left( 1+x \right) }^{ 2m }\quad$

$=$ Coefficient of

${x}^{2m}$ in

${ \left( 1-x^2 \right) }^{ 2m }$

$={ \left( -1 \right) }^{ m }\quad { _{ }^{ 2m }{ C } }_{ m }={ \left( -1 \right) }^{ n/2 }\quad { _{ }^{ n }{ C } }_{ n/2 }$

Value of

$P=\sum _{ 0\le }^{ }{ \sum _{ r<s\le n }^{ }{ { C }_{ r } } { C }_{ s } }$ is

0%
${ 2 }^{ 2n }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n } )$
0%
${ 2 }^{ 2n-1 }-\cfrac { 1 }{ 2 } ({ _{ }^{ 2n }{ C } }_{ n })$
0%
${ 2 }^{ 2n }-\cfrac 12 ({ _{ }^{ 2n }{ C } }_{ n })$
0%
None of these

11th term in the expansion of

${ \left( 3-\sqrt { \cfrac { 17 }{ 4 } +3\sqrt { 2 } } \right) }^{ 20 }$ is

0%
an irrational number
0%
a rational number
0%
a positive integer
0%
a negative integer

Explanation

$\dfrac{17}{4}+3\sqrt{2}$

$=\dfrac{1}{4}(9+8+12\sqrt{2})$

$=\dfrac{1}{4}(3+2\sqrt{2})^{2}$
Now

$(3-\sqrt{\dfrac{1}{4}(3+2\sqrt{2})^{2}})^{20}$

$=(3-\dfrac{1}{2}(3+2\sqrt{2}))^{20}$

$=(\dfrac{3}{2}-\sqrt{2})^{20}$
Writing the general term, we get

$T_{r+1}=(-1)^{r}\:^{20}C_{r}(\dfrac{3}{2})^{20-r}(\sqrt{2})^{r}$
Replacing r by 10, we get

$T_{10+1}$

$=T_{11}$

$=(-1)^{10}\:^{20}C_{r}(\dfrac{3}{2})^{20-10}(\sqrt{2})^{10}$

$=\:^{20}C_{r}.3^{10}.2^{-5}$
Hence a rational term.

$n$ is even, then value of the expression

${ C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }^{ 2 }+\cfrac { 1 }{ 3 } { C }_{ 2 }^{ 2 }-.....+\cfrac { { \left( -1 \right) }^{ n } }{ n+1 } { C }_{ n }^{ 2 }$
where

${ C }_{ r }={ _{ }^{ n }{ C } }_{ r }$ is

0%
$\cfrac { { \left( -1 \right) }^{ n }n! }{ (n+1){ (n/2)! }^{ 2 } }$
0%
$\cfrac { { \left( -1 \right) }^{ n-1 }n! }{ (n+1){ (n/2)! }^{ 2 } }$
0%
$\cfrac { { \left( -1 \right) } }{ (n+1){ (n/2)! }^{ 2 } }$
0%
$\cfrac { { \left( -1 \right) }^{ n/2 }n! }{ (n+1){ (n/2)! }^{ 2 } }$

Explanation

We have

${ C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }x+\cfrac { 1 }{ 3 } { C }_{ 2 }{ x }^{ 2 }-......+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }{ x }^{ n }\quad \cfrac { 1 }{ (n+1)x } \left[ 1-{ (1-x) }^{ n+1 } \right] ........(1)$
and

${ C }_{ 0 }+\quad { C }_{ 1 }\cfrac { 1 }{ x } +{ C }_{ 2 }\cfrac { 1 }{ { x }^{ 2 } } +........+{ C }_{ n }{ \left( \cfrac { 1 }{ x } \right) }^{ n }={ (1+x) }^{ n }\quad .....(2)$
Note that

$S={ C }_{ 0 }^{ 2 }-\cfrac { 1 }{ 2 } { C }_{ 1 }^{ 2 }+\cfrac { 1 }{ 3 } { C }_{ 2 }^{ 2 }-........+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }^{ 2 }\quad$

$=$ Coefficient of constant term in

$\left[ { C }_{ 0 }-\cfrac { 1 }{ 2 } { C }_{ 1 }x+\cfrac { 1 }{ 3 } { C }_{ 2 }{ x }^{ 2 }-......+{ \left( -1 \right) }^{ n }\cfrac { 1 }{ n+1 } { C }_{ n }{ x }^{ n } \right] \times \left[ { C }_{ 0 }+\quad { C }_{ 1 }\cfrac { 1 }{ x } +{ C }_{ 2 }\cfrac { 1 }{ { x }^{ 2 } } +........+{ C }_{ n }{ \left( \cfrac { 1 }{ { x }^{ n } } \right) } \right]$

$=$ coefficient of constant term in

$=\cfrac { 1 }{ n+1 } \left[ \cfrac { 1-{ (1-x) }^{ n+1 } }{ x } \right] { \left( 1+\cfrac { 1 }{ x } \right) }^{ n }\quad$

$=$ coefficient of

${c}^{n+1}$ in

$\cfrac { 1 }{ n+1 } \left[ { (1+x) }^{ n }-{ (1+{ x }^{ 2 }) }^{ n }(1-x) \right]$

$=-\cfrac { 1 }{ n+1 } \quad coefficient\quad of\quad { x }^{ n+1 }\quad in\quad { (1+{ x }^{ 2 }) }^{ n }(1-x)$
As

$n$ is even, let

$n=2m$ , then

$S=\cfrac { 1 }{ 2m+1 } \quad coefficient\quad of\quad { x }^{ 2m }\quad in\quad { (1+{ x }^{ 2 }) }^{ 2m }\quad$

$=\cfrac { 1 }{ 2m+1 } { \left( -1 \right) }^{ m }\left( { _{ }^{ 2m }{ C } }_{ m } \right)$

$=\cfrac { { \left( -1 \right) }^{ n } }{ n+1 } ({ _{ }^{ n }{ C } }_{ n/2 })\quad =\cfrac { { \left( -1 \right) }^{ n/2 }n! }{ (n+1){ (n/2)! }^{ 2 } }$

Let

$S={ C }_{ 1 }-\left( 1+\cfrac { 1 }{ 2 } \right) { C }_{ 2 }+\left( 1+\cfrac { 1 }{ 2 } +\cfrac { 1 }{ 3 } \right) { C }_{ 3 }-........+.{ \left( -1 \right) }^{ n-1 }\left( 1+\cfrac { 1 }{ 2 } +....+\cfrac { 1 }{ n } \right) { C }_{ n }$
then

0%
$nS=1$
0%
$\dfrac 1S$ is an integer
0%
$\dfrac 1{{ S }^{ 2 }}$ is an integer
0%
$S$ is independent of $n$

Explanation

We can write

$S$ as

$S={ a }_{ 1 }+\cfrac { 1 }{ 2 } { { a }_{ 2 } }+\cfrac { 1 }{ 3 } { a }_{ 3 }+.....+\cfrac { 1 }{ n } { a }_{ n }\quad$
where

${ a }_{ r }={ \left( -1 \right) }^{ r-1 }({ C }_{ r }-{ C }_{ r-1 }+.....+.{ \left( -1 \right) }^{ n-r }{ C }_{ n })\quad \quad ={ \left( -1 \right) }^{ n-1 }({ C }_{ 0 }-{ C }_{ 1 }+.....+.{ \left( -1 \right) }^{ n-r }{ C }_{ n-r })\quad \\ \left[ use\quad { C }_{ r }={ C }_{ n-r }\quad \right]$
But for

$k\ge 0$

${ C }_{ 0 }-{ C }_{ 1 }+{ C }_{ 2 }-.....+{ \left( -1 \right) }^{ k }{ C }_{ k }$

$=$ Coefficient of

${x}^{k}$ in

$({ C }_{ 0 }-{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }-.....+{ \left( -1 \right) }^{ n }{ C }_{ n }{ x }^{ n })\quad (1+x+{ x }^{ 2 }+....)$

$=$ Coefficient of

${x}^{k}$ in

${ \left( 1-x \right) }^{ n }{ \left( 1-x \right) }^{ -1 }$

$=$ coefficient of

${x}^{k}$ in

${ \left( 1-x \right) }^{ n-1 }$

$={ \left( -1 \right) }^{ k }({ _{ }^{ n-1 }{ C } }_{ k })$
Thus,

${ a }_{ r }={ \left( -1 \right) }^{ n-1 }{ \left( -1 \right) }^{ n-r }({ _{ }^{ n-1 }{ C } }_{ n-r })\quad$

$={ \left( -1 \right) }^{ r-1 }\quad ({ _{ }^{ n-1 }{ C } }_{ n-r })$
Now

$S={ _{ }^{ n-1 }{ C } }_{ 0 }-\cfrac { 1 }{ 2 } ({ _{ }^{ n-1 }{ C } }_{ 1 })+\cfrac { 1 }{ 3 } ({ _{ }^{ n-1 }{ C } }_{ 2 })-.....+\cfrac { { \left( -1 \right) }^{ n-1 } }{ n } ({ _{ }^{ n-1 }{ C } }_{ n-1 })$

$=\int _{ 0 }^{ 1 }{ { \left( 1-x \right) }^{ n-1 } } dx\quad \quad =\cfrac { 1 }{ n }$

values of

$x$ for which the sixth term of the expansion of

$E={ \left( { 3 }^{ \log _{ 3 }{ \sqrt { { 9 }^{ \left| x-2 \right| } } } }+{ 7 }^{ (\tfrac 15)\log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right| }-9 \right] } } \right) }^{ 7 }$ is

$567$ , are

0%
$1$
0%
$2$
0%
$3$
0%
none of these

Explanation

Put $y={ 3 }^{ \log _{ 3 }{ \sqrt { { 9 }^{ \left| x-2 \right| } } } }$ $\Rightarrow$ $\log _{ 3 }{ y } =\log _{ 3 }{ \sqrt { { 9 }^{ \left| x-2 \right| } } }$
$\Rightarrow \quad y=\sqrt { { 9 }^{ \left| x-2 \right| } } ={ 3 }^{ \left| x-2 \right| }$
Now, put $z={ 7 }^{ (\tfrac 15)\log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right| }-9 \right] } }$
$\log _{ 7 }{ z } =\cfrac { 1 }{ 5 } \log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right| }-9 \right] }$
$={ \left( \log _{ 7 }{ \left[ (4).{ 3 }^{ \left| x-2 \right| }-9 \right] } \right) }^{ \tfrac 15 }$
$\Rightarrow z={ \left[ (4).{ 3 }^{ \left| x-2 \right| }-9 \right] }^{ \tfrac 15 }$
Now, $E={ \left( y+z \right) }^{ 7 }$ and 6th term is given by
${ t }_{ 6 }={ _{ }^{ 7 }{ C } }_{ 5 }{ y }^{ 7-5 }{ z }^{ 5 }=21{ \left( { 3 }^{ \left| x-2 \right| } \right) }^{ 2 }\left\{ (4).{ 3 }^{ \left| x-2 \right| }-9 \right\}$
$\Rightarrow 567=21{ \left( { 3 }^{ 2\left| x-2 \right| } \right) }\left\{ (4).{ 3 }^{ \left| x-2 \right| }-9 \right\}$
$\Rightarrow 27=\left[ (4){ 3 }^{ 3\left| x-2 \right| } \right] -\left[ (9)({ 3 }^{ 2\left| x-2 \right| }) \right]$
$\Rightarrow \quad 27=(4){ 3 }^{ 3\left| x-2 \right| }-(9){ 3 }^{ 2\left| x-2 \right| }$
$\Rightarrow \quad 4{ u }^{ 3 }-9{ u }^{ 2 }-27=0\quad ;\quad u={ 3 }^{ \left| x-2 \right| }$
note that $u=3$ satisfies this equation
${ 3 }^{ \left| x-2 \right| }=3\quad \Rightarrow \left| x-2 \right| =1\quad \Rightarrow \quad x-2=\pm 1\quad$
$x=2\pm1=3\quad or\quad 1$ .

Sum of the series

$\sum _{ k=0 }^{ n }{ \sum _{ r=0 }^{ n-k }{ \begin{pmatrix} n \\ k \end{pmatrix} } } \begin{pmatrix} n-k \\ r \end{pmatrix}$ is

0%
${ 2 }^{ n }$
0%
${ 3 }^{ n }$
0%
$\sum _{ r=0 }^{ n }{ { (-1) }^{ r } } { C }_{ r }{ 4 }^{ r }$
0%
$\sum _{ r=0 }^{ n }{ { _{ }^{ n }{ C } }_{ r } } { 2 }^{ r }$

Explanation

$\displaystyle \sum_{k=0} ^{k=n} \sum_{r=0} ^{n-k} \:^{n}C_{k} \:^{n-k}C_{r}$
Now

$\displaystyle \sum _{0} ^{n-k} \:^{n-k}C_{r}$

$=2^{n-r}$
Therefore

$\displaystyle \sum_{k=0} ^{k=n} \:^{n}C_{k} 2^{n-k}$

$=2^{n}\:^{n}C_{0}+2^{n-1}\:^{n}C_{1}+...2^{0}\:^{n}C_{n}$

$=(2+1)^{n}$

$=3^{n}$
This can re-written as

$=(4-1)^{n}$

$\displaystyle =\sum _{r=0} ^{r=n} (-1)^{r} 4^{n-r}$

If in the expansion of

${ \left( { x }^{ 3 }-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ n }$ ,

$n\in N$ , sum of coefficient of

${ x }^{ 5 }$ and

${ x }^{ 10 }$ is

$0$ , then value of

$n$ is

0%
$5$
0%
$10$
0%
$15$
0%
none of these

Explanation

$(r+1)$ th term in the expansion of

${ \left( { x }^{ 3 }-\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ n }$ is

${ _{ }^{ n }{ C } }_{ r }{ \left( { x }^{ 3 } \right) }^{ n-r }{ \left( -\cfrac { 1 }{ { x }^{ 2 } } \right) }^{ r }={ _{ }^{ n }{ C } }_{ r }{ (-1) }^{ r }{ x }^{ 3n-5r }\quad$
For coefficient of

${ x }^{ 5 }$ , we set

$3n-5r=5$

$\Rightarrow \quad r=\cfrac { 3n-5 }{ 5 }=p \quad$ (say) and for coefficient of

${ x }^{ 10 }$ , we set

$3n-5r=10$

$\Rightarrow \quad r=\cfrac { 3n-10 }{ 5 } =q$ (say)
Note that

$p-q=1$ . We are given

${ _{ }^{ n }{ C } }_{ p }{ (-1) }^{ p }+{ _{ }^{ n }{ C } }_{ q }{ (-1) }^{ q }=0$

$\Rightarrow { _{ }^{ n }{ C } }_{ p }{ (-1) }^{ p }+{ _{ }^{ n }{ C } }_{ p-1 }{ (-1) }^{ p-1 }=0$

$\Rightarrow { _{ }^{ n }{ C } }_{ p }={ _{ }^{ n }{ C } }_{ p-1 }\quad \Rightarrow \quad n-p=p-1$

$\Rightarrow n=2p-1\quad \Rightarrow \quad p=\cfrac { n+1 }{ 2 }$
Thus,

$\cfrac { n+1 }{ 2 } =\cfrac { 3n-5 }{ 5 } \quad \Rightarrow 5n+5=6n-10\quad$

$\Rightarrow 15=n$

Value of

$S={ _{ }^{ n }{ C } }_{ r }+3({ _{ }^{ n-1 }{ C } }_{ r })+5({ _{ }^{ n-2 }{ C } }_{ r })+...+$ upto

$\quad (n-r+1)\quad terms$

0%
${ _{ }^{ n+2 }{ C } }_{ r+2 }$
0%
${ _{ }^{ n+2 }{ C } }_{ r+2 }+{ _{ }^{ n+1 }{ C } }_{ r+2 }$
0%
${ _{ }^{ n+2 }{ C } }_{ r+1 }$
0%
${ _{ }^{ n+2 }{ C } }_{ r+2 }+{ _{ }^{ n+1 }{ C } }_{ r}$

Explanation

$S=$ coefficient of

${x}^{r}$ in

$E$ where

$E={ (1+x) }^{ n }+3{ (1+x) }^{ n-1 }+5{ (1+x) }^{ n-2 }+...+(2(n-r)+1)({ (1+x) }^{ r }$
Let

$F={ a }^{ n }+3{ a }^{ n-1 }+.....(2(n-r)+1){ a }^{ r }$

$\Rightarrow \cfrac { 1 }{ a } F={ a }^{ n-1 }+....+(2(n-r)-3){ a }^{ r }+(2(n-r)+1){ a }^{ r-1 }\quad$

$\Rightarrow \left( 1-\cfrac { 1 }{ a } \right) F={ a }^{ n }+2{ a }^{ n-1 }+....+2{ a }^{ r }-(2(n-r)+1){ a }^{ r-1 }\quad$

$={ a }^{ n }+\cfrac { 2{ a }^{ r }(1-{ a }^{ n-r }) }{ 1-a } -(2(n-r)+1){ a }^{ r-1 }$

$\Rightarrow F=\cfrac { { a }^{ n+1 } }{ a-1 } -\cfrac { 2{ a }^{ r+1 }(1-{ a }^{ n-r }) }{ { (1-a) }^{ 2 } } -\cfrac { (2(n-r)+1){ a }^{ r } }{ a-1 }$
Thus,

$\quad E=\cfrac { { (1+x) }^{ n+1 } }{ x } +\cfrac { 2{ (1+x) }^{ n+1 } }{ { x }^{ 2 } } -\cfrac { 2{ (1+x) }^{ r+1 } }{ { x }^{ 2 } } -\cfrac { (2(n-r)+1){ (1+x) }^{ r } }{ x }$

$\therefore \quad S={ _{ }^{ n+1 }{ C } }_{ r+1 }+2({ _{ }^{ n+1 }{ C } }_{ r+2 })\quad ={ _{ }^{ n+2 }{ C } }_{ r+2 }+({ _{ }^{ n+1 }{ C } }_{ r+2 })$
Hence, option B.

${ S }_{ n }=1+q+{ q }^{ 2 }+{ q }^{ 3 }+...+{ q }^{ n }$ and

$\displaystyle { S' }_{ n }=1+\left( \frac { q+1 }{ 2 } \right) +{ \left( \frac { q+1 }{ 2 } \right) }^{ 2 }+...+{ \left( \frac { q+1 }{ 2 } \right) }^{ n },q\neq 1$ then

$^{ n+1 }{ { C }_{ 1 } }+^{ n+1 }{ { C }_{ 2 } }.{ S }_{ 1 }+^{ n+1 }{ { C }_{ 3 } }.{ S }_{ 2 }+...+^{ n+1 }{ { C }_{ n+1 } }.{ S }_{ n }=$

0%
${ 2 }^{ n-1 }.{ S' }_{ n }$
0%
${ 2 }^{ n }.{ S' }_{ n }$
0%
${ 2 }^{ n+1 }.{ S' }_{ n }$
0%
None of these

Explanation

$S_n$ =

$\dfrac{q^{n+1}-1}{q-1}$ summation of g.p

$^{n+1}{C}_{1}$ +

$^{n+1}{C}_{2}$

$\dfrac{q^2-1}{q-1}$ +

$^{n+1}{C}_{3}$

$\dfrac{q^3-1}{q-1}$ +..................+

$^{n+1}{C}_{n+1}$

$\dfrac{q^{n+1}-1 }{q-1}$

$\dfrac{^{n+1}{C}_{0}}{q-1}$ -

$^{n+1}{C}_{1}$

$\dfrac{q-1}{q-1}$ +

$^{n+1}{C}_{2}$

$\dfrac{q^2-1}{q-1}$ +

$^{n+1}{C}_{3}$

$\dfrac{q^3-1}{q-1}$ +..................+

$^{n+1}{C}_{n+1}$

$\dfrac{q^{n+1}-1 }{q-1}$ -

$\dfrac{^{n+1}{C}_{0}}{n-1}$

(

$\dfrac{^{n+1}{C}_{0}}{q-1}$ +

$^{n+1}{C}_{1}$

$\dfrac{q}{q-1}$ +

$^{n+1}{C}_{2}$

$\dfrac{q^2}{q-1}$ +

$^{n+1}{C}_{3}$

$\dfrac{q^3}{q-1}$ +..................+

$^{n+1}{C}_{n+1}$

$\dfrac{q^{n+1} }{q-1}$ )-(

$\dfrac{^{n+1}{C}_{0}}{q-1}$ +

$^{n+1}{C}_{1}$

$\dfrac{}{q-1}$ +

$^{n+1}{C}_{2}$

$\dfrac{}{q-1}$ +

$^{n+1}{C}_{3}$

$\dfrac{}{q-1}$ +.................. +

$^{n+1}{C}_{n+1}$

$\dfrac{}{q-1}$ )

$\dfrac{{(1+q)}^{n+1}}{q-1}$ -

$\dfrac{2^{n+1}}{q-1}$

$2^n(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$ -

$\dfrac{2}{q-1})$

$2^n(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$ -

$\dfrac{1}{\dfrac{q-1}{2}})$

$2^n$

$S'_n$

and

$S'_n$ =

$(\dfrac{{(\dfrac{1+q}{2}})^{n+1}}{\dfrac{q-1}{2}}$

$-$

$\dfrac{1}{\dfrac{q-1}{2}})$ summation of g.p

The third term from the end in the expansion of

$\displaystyle\left(\frac{4x}{3y}-\frac{3y}{2x}\right)^9$ is

0%
$\displaystyle^9C_7\frac{3^5}{2^3}\frac{y^5}{x^5}$
0%
$\displaystyle^{-9}C_7\frac{3^5}{2^3}\frac{y^5}{x^5}$
0%
$\displaystyle^9C_7\frac{3^5}{2^3}\frac{y^5}{x^3}$
0%
none of these

Explanation

Given,

$\left(\dfrac{4x}{3y}-\dfrac{3y}{2x}\right)^2$

The third term from the end in this expansion is

$T_3=T_{2+1}={^{9}C_2}\left(-\dfrac{3y}{2x}\right)^7\left(\dfrac{4x}{3y}\right)^2$

$T_3=-{^{9}C_2}\dfrac{3^5}{2^3}\dfrac{y^5}{x^5}$

Trick: Remember whenever asked the term from end just reverse 'x' and 'y' in the general term formula given by,

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

i.e., for end term

$T_{r+1}={^{n}C_r}y^{n-r}x^r$ .

If the second ,third and fourth terms in the expansion of

${\left(x+y\right)}^{n}$ are

$240,\,720$ and

$1080$ respectively, then the value of

$x,\,y,\,n$ is

0%

$x=2,\,y=3,\,n=5$
0%

$x=3,\,y=3,\,n=5$
0%

$x=2,\,y=3,\,n=3$
0%

$x=2,\,y=2,\,n=5$

Maximum sum of the coefficients in the expansion of

$(1 - x \sin \theta+ x^{2} )^{n}$ is

0%
$1$
0%
$2^{n}$
0%
$3^{n}$
0%
$0$

Explanation

${ \left( 1-x\sin\theta +{ x }^{ 2 } \right) }^{ n }$

We can get the sum of the coefficient by directly taking

$x=1$ in the above expression.

$\Rightarrow S={ \left( 1-\sin\theta +1 \right) }^{ n }$

$={ \left( 2-\sin\theta \right) }^{ n }$

$-1\le\sin\theta\le1$

$\Rightarrow$ Minimum value of

$\sin\theta=-1$

$\therefore$

$S_{max}={(2-(-1))}^n$

$=3^n.$

Hence, the answer is

$3^n.$

$C_{0}+3.C_{1}+3.^{2}\textrm{C}_{2}+...+3.^{n}C_{n}=5^{n}.$

0%
True
0%
False

Explanation

$\left ( 1+x \right )^{n}=C_{0}+C_{1}x+C_{2}x^{2}+...+C_{n}x^{n}$
Put

$x=3$

$C_{0}+3.C_{1}+3^{2}C_{2}+...+3^{n}C_{n}=4^{n}$

$\left( _{ }^{ m }{ { C }_{ 0 }^{ } }+^{ m }{ { C }_{ 1 }^{ } }-^{ m }{ { C }_{ 2 }^{ } }-^{ m }{ { C }_{ 3 }^{ } } \right) +\left( ^{ m }{ { C }_{ 4 }^{ } }+^{ m }{ { C }_{ 5 }^{ } }-^{ m }{ { C }_{ 6 }^{ } }-^{ m }{ { C }_{ 7 }^{ } } \right) +...=0$ if and only if for some positive integer

$k, m=$

0%
$4k$
0%
$4k+1$
0%
$4k-1$
0%
$4k+2$

Explanation

Consider

${ \left( \cos { \theta } -i\sin { \theta } \right) }^{ m }=^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } - }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } +...+^{ m }{ { C }_{ m } }{ \left( -i\sin { \theta } \right) }^{ m }$ ....(1)

${ \left( \cos { \theta } +i\sin { \theta } \right) }^{ m }=^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } +...+^{ m }{ { C }_{ m } }{ \left( i\sin { \theta } \right) }^{ m }$ ....(2)

Adding (1) and (2), we get

$2\cos { m\theta =2\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta ... } } } \right] }$ ...(3)

Subtracting (3) and (4), we get

$2i\sin { m\theta } =2i\left[ ^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right]$ ....(4)

Adding (3) and (4), we get

$\cos { m\theta } +\sin { m\theta } \\ =\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta - } }^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right]$

$\displaystyle \Rightarrow \sqrt { 2 } \sin { \left( m\theta +\frac { \pi }{ 4 } \right) } \\ =\left[ ^{ m }{ { C }_{ 0 }\cos ^{ m }{ \theta } + }^{ m }{ { C }_{ 1 }\cos ^{ m-1 }{ \theta } i }\sin { \theta } -^{ m }{ { C }_{ 2 }\cos ^{ m-2 }{ \theta } \sin ^{ 2 }{ \theta - } }^{ m }{ { C }_{ 3 }\cos ^{ m-3 }{ \theta } \sin ^{ 3 }{ \theta ... } } \right]$

Putting

$\displaystyle \theta =\frac { \pi }{ 4 }$ , we get

$\displaystyle \sqrt { 2 } \sin { \left( \frac { \left( m+1 \right) \pi }{ 4 } \right) } =\frac { 1 }{ { 2 }^{ \frac { m }{ 2 } } } [\left( ^{ m }{ { C }_{ 0 } }+^{ m }{ { C }_{ 1 }-^{ m }{ { C }_{ 2 }-^{ m }{ { C }_{ 3 } } } } \right) \\ +\left( ^{ m }{ { C }_{ 4 }+^{ m }{ { C }_{ 5 }-^{ m }{ { C }_{ 6 } }-^{ m }{ { C }_{ 7 } } } } \right) +...+\left( ^{ m }{ { C }_{ m-3 }+^{ m }{ { C }_{ m-2 }-^{ m }{ { C }_{ m-1 }-^{ m }{ { C }_{ m } } } } } \right) ]$

Hence

$m+1=4k$ , for given quantity to be

$0$ .

$\Rightarrow m=4k-1$ where

$k\in N$

In the expansion of

$\left (5^{ \tfrac {1}{2}}+7^{\tfrac {1}{8}}\right )^{1024}$ , the number of integral terms is

0%
$128$
0%
$129$
0%
$130$
0%
$131$

Explanation

The general term in the expansion of

$(5^{\tfrac 12} + 7^{\tfrac 18})^{1024}$ is

$T_{r+1} = ^{1024}C_r (5^{\tfrac 12})^{1024-r} (7^{\tfrac 18})^r$

$= ^{1024}C_r 5^{512-\tfrac{r}{2}} 7^{\tfrac{r}{8}}$

$= (^{1024}C_r 5^{512-r})(5^{\tfrac{r}{2}}7^{\tfrac{r}{8} })$

$= (^{1024}C_r 5^{512r}) (5^4\times 7)^{\tfrac{r}{8}}$

Clearly

$T_{r+1}$ will be an integer, if

$\dfrac{r}{8}$ is an integer such that

$0\leq r\leq 1024$

$\Rightarrow r$ is a multiple of

$8$ lying satisfying

$0\leq r \leq 1024$

$r = 0, 8, 16, ... 1024$

$\Rightarrow r$ can assume

$129$ values.
Hence, there are

$129$ integral terms in the expansion of

$(5^{\tfrac 12} + 7^{\tfrac 18})^{1024}$

If the expansion of

$\displaystyle\left(x^3+\frac{1}{x^2}\right)^n$ contains a term independent of x, then the value of n can be

0%
18
0%
20
0%
24
0%
22

Explanation

Given,

$\left(x^3+\dfrac{1}{x^2}\right)^n$

this contains a term independent of x.

so general term

$T_{r+1}$

$T_{r+1}={^{n}C_r}x^{n-r}y^r\\$

$T_{r+1}={^{n}C_r}(x^3)^{n-r}\left(\dfrac{1}{x^2}\right)^r\\$

$T_{r+1}={^{n}C_r}\dfrac{(x)^{3n-3r}}{x^{2r}}\\$

$T_{r+1}={^{n}C_r}(x)^{3n-5r}$

Since, given term is indepedent of x.

so,

$3n-5r=0$

$3n=5r$

$n=\dfrac{5r}{3}$

Since n should be an integer, using options

$n=9$

$n=\dfrac{5\times 9}{3}$

$n=15$ (not in options)

so,

$r=15$ ,

$n=\dfrac{5\times 15}{3}$ ,

$n=25$ (not in options)

so,

$r=12$ ,

$n=\dfrac{5\times 12}{3}$ ,

$n=20$ (in options).

In the expansion of

${ \left( \dfrac { 3{ x }^{ 2 } }{ 5 } +\dfrac { 5 }{ 3{ x }^{ 2 } } \right) }^{ 10 }$ mid term is

0%
$291$
0%
$242$
0%
$252$
0%
$284$

Explanation

Given expansion

$={ \left( \dfrac { 3{ x }^{ 2 } }{ 5 } +\dfrac { 5 }{ 3{ x }^{ 2 } } \right) }^{ 10 }$
On comparing with

${ \left( x+a \right) }^{ n }$ , we get

$x=\dfrac { 3{ x }^{ 2 } }{ 5 }$ ,

$a=\dfrac { 5 }{ 3{ x }^{ 2 } }$ and

$n=10$

$\because$ Number of terms

$=10+1=11$

$\therefore$ Mid term

$=\left( \dfrac { 11+1 }{ 2 } \right)$ th term

$=6$ th term
Now,

${ T }_{ 6 }={ T }_{ 5+1 }=_{ }^{ 10 }{ { C }_{ 5 } }{ \left[ \dfrac { 3{ x }^{ 2 } }{ 5 } \right] }^{ 10-5 }{ \left[ \dfrac { 5 }{ 3{ x }^{ 2 } } \right] }^{ 5 }$

$\left[ \because { T }_{ r+1 }=_{ }^{ n }{ { C }_{ r } }{ x }^{ n-r }{ a }^{ r } \right]$

$=_{ }^{ 10 }{ { C }_{ 5 } }{ \left[ \dfrac { 3{ x }^{ 2 } }{ 5 } \right] }^{ 5 }{ \left[ \dfrac { 5 }{ 3{ x }^{ 2 } } \right] }^{ 5 }=_{ }^{ 10 }{ { C }_{ 5 } }=252$ .

$(1+x)^{2n} =a_0+a_1x....+a_{2n}x^{2n}$ , then

0%
$a_1+a_2+a_4.....=\dfrac 12 (a_0+a_1+a_2.....)$
0%
$a_{n+1}=a_n$
0%
$a_{n-3}=a_{n+3}$
0%
$a_{n-3}>a_{n+3}$

Explanation

${ \left( 1+x \right) }^{ 2n }+{ \left( 1-x \right) }^{ 2n }=2\left( { a }_{ 0 }+{ a }_{ 2 }{ x }^{ 2 }+{ a }_{ 4 }{ x }^{ 4 }+...+{ a }_{ 2n }{ x }^{ 2n } \right)$

For

$x=1$ ,

$\dfrac { { 2 }^{ 2n } }{ 2 } =\left( { a }_{ 0 }+{ a }_{ 2 }+{ a }_{ 4 }+...+{ a }_{ 2n } \right)$

${ \left( 1+x \right) }^{ 2n }=\left( { a }_{ 0 }+{ a }_{ 1 }x+{ a }_{ 2 }{ x }^{ 2 }+...+{ a }_{ 2n }{ x }^{ 2n } \right)$ ...

$(i)$

For

$x=1$ ,

$2^{2n}=a_0+a_1+a_2+....+a_{2n}$

Hence,

$\left( { a }_{ 0 }+{ a }_{ 2 }+{ a }_{ 4 }+...+{ a }_{ 2n } \right) =$

$1/2(a_0+a_1+a_2+....+a_{2n})$

So,

$A$ is correct.

Now,

$\dfrac {(1+x)^{2n}}{x^{2n}}=\dfrac{(a_0+a_1x+...+a_{2n}x^{2n})}{x^{2n}}$

So,

${ \left( 1+\dfrac { 1 }{ x } \right) }^{ 2n }=\dfrac { { a }_{ o } }{ { x }^{ 2n } } +\dfrac { { a }_{ 1 } }{ { x }^{ 2n-1 } } +....+\dfrac { { a }_{ 2n } }{ { x }^{ 0 } }$ ...

$(ii)$

Putting

$x=\dfrac{1}{x}$ in

$(i)$ ,

${ \left( 1+\dfrac { 1 }{ x } \right) }^{ 2n }=a_0+\dfrac{a_1}{x}+\dfrac{a_2}{x^{2}}+...+\dfrac{a_{2n}}{x^{2n}}$ ...

$(iii)$

On comparing coefficients from

$(ii)$ and

$(iii)$ ,

$a_{2n}=a_0$ ,

$a_{2n-1}=a_1$ .... and so on...

So, we can say,

$a_{2n-r}=a_r$ . for

$2n \ge r \ge 0$

For,

$r=n+3$ ,

$a_{n-3}=a_{n+3}$ So,

$C$ is correct.

But for

$r=n$ , we find

$B$ is incorrect.

$ac>b^2$ then the sum of the coefficients in the expansion of

$(a\alpha ^2x^2+2b\alpha x+c)^n,(a,b,c,\alpha \in R, n\in N)$ is

0%
Positive if $a>0$ .
0%
Positive if $c>0$ .
0%
Negative if $a<0, n$ is odd.
0%
Positive if $c<0,n$ is even.

Explanation

$ac>b^2$ , then

$ac-b^2>0$ or

$b^2-ac<0$

Now,

${ \quad \left( a{ \alpha }^{ 2 }{ x }^{ 2 }+2b\alpha x+c \right) }^{ n }\\ ={ \left( a{ \left( \alpha x \right) }^{ 2 }+2b\left( \alpha x \right) +c \right) }^{ n }$

Here,

$\Delta =4{ b }^{ 2 }-4ac\\ \quad =4\left( { b }^{ 2 }-ac \right) \\ \quad <0$

Hence, it has no roots.

the point where slope is 0 in

$a{ x }^{ 2 }+bx+c$ is

$y=\dfrac { -\left( { b }^{ 2 }-4ac \right) }{ 4a }$

so for

$a>0$ , y=+ve as (

$\Delta <0$ )

$a<0$ , y=-ve

hence sum of coefficient is +ve for

$a>0$

So, correct option is

$A$ and

$C$ .

If the sum of the coefficients in the expansion of

$(l^2x^2-2lx+1)^{50}$ vanishes then

$l$ is equal to:

0%
$-1$
0%
$-2$
0%
$1$
0%
$2$

Explanation

${(l^2x^2-2lx+1)}^{50}$ =

${(lx-1})^{100}$

If the sum of the coefficcents of the expansion vanishes mens equal to 0

means on putting x=1

$\Rightarrow$

$lx-1=0$

$\Rightarrow$

$l-1=0$

$\Rightarrow$

$l=1$

Find the value(s) of k such that the term independent of x in

$\displaystyle\left(3x^2+\frac{k}{2x}\right)^6$ is 135.

0%
$\pm2$
0%
$\pm1$
0%
$\pm3$
0%
$\pm4$

Explanation

Given,

$\left(3x^2+\dfrac{k}{2x}\right)^6$

general term,

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{6}C_r}(3x^2)^{6-r}\left(\dfrac{k}{2x}\right)^r$

$T_{r+1}={^{6}C_r}(3)^{6-r}(x)^{12-2r}\dfrac{(k)^r}{(2)^rx^r}$

$T_{r+1}=\dfrac{^{6}C_r(3)^{6-r}(k)^r}{(2)^r}(x)^{12-3r}$

given term is independent of

$x$

$12-3r=0$

$12=3r$

$r=4$

given term independent of x is

$135$

$135={^{6}C_4}\dfrac{(3)^{6-4}k^4}{(2)^4}$

$135=\dfrac{6!}{4!\times 2!}\dfrac{(3)^2}{(2)^4}k^4$

$135=\dfrac{5\times 6(3)^2}{2(2)^4}k^4$

$k^4=\dfrac{135\times 2\times 2}{5\times 6\times 3\times 3}$

$k^4=2^4$

$k=\pm 2$ .

Find the coefficient of

$x^4$ in the expansion of

$\left(2x^2+\frac{3}{x^3}\right)^7$

0%
$^7C_22^53^3$
0%
$^7C_22^53^2$
0%
$^7C_23^52^2$
0%
$^7C_32^53^2$

Explanation

Given,

$\left(2x^2+\dfrac{3}{x^3}\right)^7$

To find coefficient of

$x^4$

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{7}C_r}(2x^2)^{7-r}\left(\dfrac{3}{x^3}\right)^r$

$T_{r+1}={^{7}C_r}(2)^{7-r}(x)^{14-2r}\dfrac{(3)^r}{x^{3r}}$

$T_{r+1}={^{7}C_r}(2)^{7-r}(3)^r\times (x)^{14-5r}$ ........

$(1)$

given to find coefficient of

$x^4$

$14-5r=4$

$14-4=5r$

$5r=10$

$r=2$

so putting

$r=2$ in

$(1)$

$T_{r+1}={^{7}C_2}(2)^{7-2}(3)^2(x)^4$

$T_{r+1}={^{7}C_2}2^53^2x^4$

$\therefore$ So, coefficient of

$x^4$ in the expansion of

$\left(2x^2+\dfrac{3}{x^3}\right)^7$ is

$^{7}C_22^53^2$ .

The sum of the series

$\frac{1}{1\times 2}^{25}C_0 + \frac{1}{2\times 3}^{23}C_1+\frac{1}{3\times 4}^{25}C_2+...... + \frac{1}{26\times 27}^{25}C_{25}$

0%
$\dfrac{2^{27}-1}{26\times 27}$
0%
$\dfrac{2^{27}-28}{26\times 27}$
0%
$\dfrac{1}{2}\left(\frac{2^{26}+1}{26\times 27}\right)$
0%
$\dfrac{2^{26}-1}{52}$

Explanation

$(1+x)^{25}=\displaystyle^{25}C_{0}+\displaystyle^{25}C_{1}x+\displaystyle^{25}C_{2}x^2+....+\displaystyle^{25}C_{25}x^{25}-------(1)$

Intergrating eq (1) w.r.t

$x$ with limit

$0$ to

$x$

$\int_{0}^{x}(1+x)^{25}=\int_{0}^{x}(\displaystyle^{25}C_{0}+\displaystyle^{25}C_{1}x+\displaystyle^{25}C_{2}x^2+....+\displaystyle^{25}C_{25}x^{25})$

$\Rightarrow \dfrac{(1+x)^{26}-1}{26}=\displaystyle^{25}C_{0}x+\dfrac{\displaystyle^{25}C_{1}x^2}{2}+\dfrac{\displaystyle^{25}C_{2}x^3}{3}+....+\dfrac{\displaystyle^{25}C_{25}x^{26}}{26}----(2)$

Intergrating eq (2) w.r.t

$x$ with limit

$0$ to

$1$ , we get

$\left [\dfrac{\displaystyle^{25}C_{0}x^2}{2}+\dfrac{\displaystyle^{25}C_{1}x^3}{2\times 3}+\dfrac{\displaystyle^{25}C_{2}x^4}{3\times 4}+....+\dfrac{\displaystyle^{25}C_{25}x^{27}}{26\times 27} \right ]^1_{0}=\left [ \dfrac{(1+x)^{27}}{26\times 27}-\dfrac{x}{26} \right ]^1_{0}$

$\dfrac{\displaystyle^{25}C_{0}}{1\times 2}+\dfrac{\displaystyle^{25}C_{1}}{2\times 3}+\dfrac{\displaystyle^{25}C_{2}}{3\times 4}+....+\dfrac{\displaystyle^{25}C_{25}}{26\times 27}=\dfrac{(2)^{27}-1}{26\times 27}-\dfrac{1}{26}$

The number of rational terms in the expansion of

$\left(x^{\displaystyle\frac{1}{5}}+y^{\displaystyle\frac{1}{10}}\right)^{45}$ is

0%
5
0%
6
0%
4
0%
7

Explanation

Given,

$\left(x^{\dfrac{1}{5}}+y^{\dfrac{1}{10}}\right)^{45}$

general term

$T_{r+1}={^{n}C_r}x^{n-r}y^r$

$T_{r+1}={^{45}C_r}(x^{\dfrac{1}{5}})^{45-r}(y^{\dfrac{1}{10}})$

$T_{r+1}={^{45}C_r}x^{9-\dfrac{r}{5}}(y)^{\dfrac{r}{10}}$

Rational term when

$r=0, 10, 20, 30, 40$

so, there are

$5$ rational numbers.

1072544_397171_ans_43ace98655784d3ebd45f1d2b943e3b1.png

The value of

$x$ in the expression

${ \left( x+{ x }^{ \log _{ 10 }{ x } } \right) }^{ 5 }$ , if the third term in the expansion is

$1,000,000$ , is

0%
$10,{ 10 }^{ { -3 }/{ 2 } }$
0%
$100$ or ${ 10 }^{ { -3 }/{ 2 } }$
0%
$10$ or ${ 10 }^{ { -5 }/{ 2 } }$
0%
None of these

Explanation

$\log { x }$ is defined only when

$x>0$
Now, the

${ 3 }^{ rd }$ term in the expansion

${ T }_{ 2+1 }=^{ 5 }{ { C }_{ 2 } }\cdot { x }^{ 5-2 }\cdot { \left( { x }^{ \log _{ 10 }{ x } } \right) }^{ 2 }=1,000,000$ ....(Given)

$\Rightarrow { x }^{ 3+2\log _{ 10 }{ x } }={ 10 }^{ 5 }$
Taking logarithm of both sides, we get

$\left( 3+2\log _{ 10 }{ x } \right) \cdot \log _{ 10 }{ x } =5$

$\Rightarrow 2{ y }^{ 2 }+3y-5=0$
where

$\log _{ 10 }{ x } =y$

$\Rightarrow \left( y-1 \right) \left( 2y+5 \right) =0$

$\Rightarrow y=1$ or

${ -5 }/{ 2 }$

$\Rightarrow \log _{ 10 }{ x } =1$ or

${ -5 }/{ 2 }$

$\Rightarrow x={ 10 }^{ 1 }=10$ or

${ 10 }^{ { -5 }/{ 2 } }$

Sum of the last

$30$ coefficients in the expansion of

${ \left( 1+x \right) }^{ 59 }$ , when expanded in ascending power of

$x$ is

0%
${ 2 }^{ 59 }$
0%
${ 2 }^{ 58 }$
0%
${ 2 }^{ 30 }$
0%
${ 2 }^{ 29 }$

Explanation

${ (1+x) }^{ n }=^{ n }C_{ 0 }+^{ n }C_{ 1 }x+^{ n }C_{ 2 }{ x }^{ 2 }.........^{ n }C_{ n }{ x }^{ n }\\ \Rightarrow ^{ n }C_{ 0 }+^{ n }C_{ 1 }+^{ n }C_{ 2 }.........^{ n }C_{ n }={ 2 }^{ n }$

For the given expansion ${ (1+x) }^{ 59 }$ let sum of last $30$ coefficients be $S$

$S=^{ 59 }C_{ 30 }+^{ 59 }C_{ 31 }+^{ 59 }C_{ 32 }..........^{ 59 }C_{ 59 }$ ....(i)

As $^{ n }C_{ r }=^{ n }C_{ n-r }$

$\Rightarrow S=^{ 59 }C_{ 29 }+^{ 59 }C_{ 28 }+^{ 59 }C_{ 27 }..........^{ 59 }C_{ 0 }$ ....(ii)

Adding (i) and (ii), we get

$\Rightarrow 2S=^{ 59 }C_{ 0 }+^{ 59 }C_{ 1 }+^{ 59 }C_{ 2 }........^{ 59 }C_{ 59 }\\ \Rightarrow 2S={ 2 }^{ 59 }\\ \Rightarrow S={ 2 }^{ 58 }$

So, option B is correct.

If there is a term containing

$x^{2r}$ in

$\left( x + \dfrac{1}{x^2} \right )^{n - 3}$ , then

0%
n - 2r is a positive integral multiple of 3.
0%
n - 2r is even
0%
n - 2r is odd
0%
None of the above

Explanation

Suppose

$(s + 1)^{th}$ term conains

$x^{2r}$
Then, we have

$T_{s + 1} = ^{n - 3}C_s x^{n - 3- s} \left( \dfrac{1}{x^2} \right )^s$

$= ^{n - 3}C_s x^{n - 3 - 3s}$
This will contain

$x^{2r}$ , if

$n - 3 - 3 s = 2r$

$\Rightarrow s = \dfrac{n - 3 - 2r}{3}$

$\Rightarrow s = \dfrac{n - 2r}{3} - 1$

$\Rightarrow s + 1 = \dfrac{n - 2r}{3}$

$\Rightarrow n - 2r = 3 ( s + 1)$
Hence,

$(n - 2r)$ is a positive integral multiple of

$3$ .

The term independent of

$x$ in the expansion of

$\left [\sqrt {\dfrac {x}{3}} + \sqrt {\dfrac {3}{2x^{2}}} \right ]^{10}$ is

0%
$1$
0%
$^{10}C_{1}$
0%
$\dfrac {5}{12}$
0%
None of these

Explanation

$T_{r + 1} = ^{10}C_{r} \left (\dfrac {x}{3}\right )^{\tfrac {10 - r}{2}} \left (\dfrac {3}{2x^{2}} \right )^{\tfrac {r}{2}}$

$= ^{10}C_{r} x^{\tfrac {10 - r}{2} - r} \left (\tfrac {1}{2}\right )^{\tfrac {r}{2}} 3^{r - 5}$ Put

$\dfrac {10 - r}{2} - r = 0$

$\Rightarrow r = \dfrac {10}{3} \not {\epsilon}$ whole number

$\therefore$ No choice match

$\therefore$ (d) is correct answer.

Coefficient of

$x^n$ in the expansion of

$\left(\displaystyle 1+\frac{x}{1!}+\frac{x^2}{2!}+...+\frac{x^n}{n!}\right)^2$ is?

0%
$\displaystyle\frac{2^n}{n!}$
0%
$\displaystyle\frac{2^{n-1}}{n!}$
0%
$\displaystyle\frac{2^{n+1}}{n!}$
0%
None of these

$\sum { { \left( -1 \right) }^{ r } } ~ { _{ }^{ n }{ C } }_{ r }\cfrac { 1+r\log _{ e }{ 10 } }{ { \left( 1+\log _{ e }{ { 10 }^{ n } } \right) }^{ r } }$

0%
$1$
0%
$-1$
0%
$n$
0%
none of these

$\sum _{ r=0 }^{ n-1 }{ { \left( \cfrac { { _{ }^{ n }{ C } }_{ r } }{ { _{ }^{ n }{ C } }_{ r }+{ _{ }^{ n }{ C } }_{ r+1 } } \right) }^{ 3 } } =\cfrac { 4 }{ 5 }$ then

$n=$

0%
$4$
0%
$6$
0%
$8$
0%
None of these

Explanation

We know that

$\left( \begin{matrix} n \\ r+1 \end{matrix} \right) +\left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right)$ ,

$\Longrightarrow \sum _{ 0 }^{ n-1 }{ { \left( \frac { \left( \begin{matrix} n \\ r \end{matrix} \right) }{ \left( \begin{matrix} n+1 \\ r+1 \end{matrix} \right) } \right) }^{ 3 } } =\sum _{ 0 }^{ n-1 }{ { \left( \frac { r+1 }{ n+1 } \right) }^{ 3 } }$

$\\ \Longrightarrow \frac { { 1 }^{ 3 }+{ 2 }^{ 3 }+{ 3 }^{ 3 }+...+{ n }^{ 3 } }{ { \left( n+1 \right) }^{ 3 } } \\ \Longrightarrow \frac { { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } }{ 4{ \left( n+1 \right) }^{ 3 } } \\ \Longrightarrow \frac { { n }^{ 2 } }{ 4{ \left( n+1 \right) } }$

Given that

$\frac { { n }^{ 2 } }{ 4{ \left( n+1 \right) } } =\frac { 4 }{ 5 }$

$\Longrightarrow 5{ n }^{ 2 }-16n-16=0\\ \Longrightarrow \left( n-4 \right) \left( 5n+4 \right) =0$

$n$ is a natural number

$\Longrightarrow n=4$

$(1+x)^{10} = a_0 + a_1x + a_2x^2 + ..... + a_{10}x^{10}$ , then value of

$(a_0 -a_2 + a_4 - a_6 + a_8 - a_{10})^2 + (a_1 -a_3 + a_5 - a_7 + a_9)^2$ is

0%
$2^{10}$
0%
$2$
0%
$2^{20}$
0%
$2^{30}$

Explanation

Here

${ a }_{ i }=^{ 10 }{ C }_{ i }$

ie,

${ ({ a }_{ 0 }-{ a }_{ 2 }+{ a }_{ 4 }-{ a }_{ 6 }+{ a }_{ 8 }-{ a }_{ 10 }) }^{ 2 }+{ ({ a }_{ 1 }-{ a }_{ 3 }+{ a }_{ 5 }-{ a }_{ 7 }+{ a }_{ 9 }) }^{ 2 }$

$={ (^{ 10 }{ C }_{ 0 }-^{ 10 }{ C }_{ 2 }+^{ 10 }{ C }_{ 4 }-^{ 10 }{ C }_{ 6 }+^{ 10 }{ C }_{ 8 }-^{ 10 }{ C }_{ 10 }) }^{ 2 }+{ (^{ 10 }{ C }_{ 1 }-^{ 10 }{ C }_{ 3 }+^{ 10 }{ C }_{ 5 }-^{ 10 }{ C }_{ 7 }+^{ 10 }{ C }_{ 9 }) }^{ 2 }$

$\Rightarrow { ((^{ 10 }{ C }_{ 0 }-^{ 10 }{ C }_{ 10 })+(^{ 10 }{ C }_{ 8 }-^{ 10 }{ C }_{ 2 })+(^{ 10 }{ C }_{ 4 }-^{ 10 }{ C }_{ 6 })) }^{ 2 }+{ ( (-1){ (-2) }^{ \frac { 10 }{ 2 } }) }^{ 2 }={ 2 }^{ 10 }$

Since

$\sum _{ i=0 }^{ \left[ n/2 \right] }{ ^{ n }{ C }_{ 2i } } =\begin{cases} 0,\quad \text{If}\quad \dfrac { n+2 }{ 4 } \in \quad \text{Integers} \\ { (-1) }^{ \left[ (n+2)/4 \right] }{ 2 }^{ \left[ n/2 \right] } \end{cases}$

and

$\sum _{ i=0 }^{ \left[ n/2 \right] }{ ^{ n }{ C }_{ 2i+1 } } =\begin{cases} 0,\quad \text{If}\quad \dfrac { n }{ 4 } \in \text{Integers} \\ { (-1) }^{ [n/4] }{ 2 }^{ [n/2] } \end{cases}$

Option A is correct

$\left\{ x \right\}$ denotes the fraction part of

$'x'$ , then

$\left\{ \dfrac { { 3 }^{ 1001 } }{ 82 } \right\} =$

0%
$\dfrac { 9 }{ 82 }$
0%
$\dfrac { 81 }{ 82 }$
0%
$\dfrac { 3 }{ 82 }$
0%
$\dfrac { 1 }{ 82 }$

Explanation

$\Rightarrow \left \{ \dfrac{3^{1001}}{82} \right \}$

$= 3^{4} \rightarrow 81$

$1-\dfrac{1}{82} -1$

$= \dfrac{(81)^{997}}{82}$

$\boxed{\dfrac{81}{82}}$

$= \dfrac{(82-1)^{997}}{82}$

$= \left \{ -\dfrac{(1-82)^{997}}{82} \right \}$

$= ^{997}C_{0} (-82)^0 + ^{997}C_{1} \dfrac{(-82)^1}{82}$

$= -\dfrac{1}{82}$

All will in proper division of 82.

$= 1-\dfrac{1}{82}-1$

$= \left \{ \dfrac{81}{82}-1 \right \}$

$= \dfrac{81}{82}$

1057730_801283_ans_f5f1aa85e2164428a724d024176e8137.jpg

The coefficient of

$x^{160}$ in the expansion of

$\displaystyle (x^8 + 1)^{60} \left( x^{12} + 3x^4 + \frac{3}{x^4} + \frac{1}{x^{12}} \right)^{-10}$ is

0%
$\displaystyle ^{30}C_6$
0%
$\displaystyle ^{30}C_5$
0%
divisible by 189
0%
divisible by 203

Explanation

$S = { \left( 1+{ x }^{ 8 } \right) }^{ 60 }.{ \left( { x }^{ 12 }+{ 3x }^{ 4 }+\dfrac { 3 }{ { x }^{ 4 } } +\dfrac { 1 }{ { x }^{ 12 } } \right) }^{ -10 }$

$={ x }^{ 12 }+{ 3x }^{ 4 }+\dfrac { 3 }{ { x }^{ 4 } } +\dfrac { 1 }{ { x }^{ 12 } } ={ \left( { x }^{ 4 }+\dfrac { 1 }{ { x }^{ 4 } } \right) }^{ 3 }$

$\Rightarrow 3 = { (1+{ x }^{ 8 }) }^{ 60 }.{ \left\{ { \left( \dfrac { { x }^{ 8 }+1 }{ { x }^{ 4 } } \right) }^{ 3 } \right\} }^{ -10 }$

$=\dfrac { { (1+{ x }^{ 8 }) }^{ 60 }.{ (1+{ x }^{ 8 }) }^{ -30 } }{ { x }^{ -120 } }$

$={ (1+{ x }^{ 8 }) }^{ 30 }.{ x }^{ 120 }$

$=[1+{ ^{ 30 }C }_{ 1 }.{ x }^{ 8 }+.....{ ^{ 30 }C }_{ 5 }.{ (x }^{ 8 })^{ 5 }].{ x }^{ 120 }$

$\therefore$ Coefficient of

${ x }^{ 160 }={ ^{ 30 }C }_{ 5 }.$

Hence the answer is

${ ^{ 30 }C }_{ 5 }.$

The value of

$\sum _{ r=1 }^{ 10 }{ \left( \sin { \cfrac { 2nr }{ 11 } } -i\cos { \cfrac { 2nr }{ 11 } } \right) }$ is

0%
$0$
0%
$-1$
0%
$-i$
0%
$i$

The co-efficient of

${x^{53}}$ in the expression

$\sum\limits_{m = 0}^{100} {{}^{100}} {c_m}{(x - 3)^{100 - m}}{2^m}\,$ is

0%
${}^{100}{c_{53}}$
0%
${}^{98}{c_{53}}$
0%
${}^{65}{c_{53}}$
0%
${}^{100}{c_{65}}$

Explanation

$S=\sum_{m=0}^{100}{^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}}$

$^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}$ is general term of

${\left(x-3+2\right)}^{100}$

Hence

$\sum_{m=0}^{100}{^{100}C_{m}{\left(x-3\right)}^{100-m}{2}^{m}}={\left(x-3+2\right)}^{100}$

${\left(x-1\right)}^{100}={\left(1-x\right)}^{100}$

Hence coefficient of

${x}^{53}$ in

${\left(1-x\right)}^{100}={\left(-1\right)}^{53}\,^{100}C_{53}=-^{100}C_{53}$

In the expression of

$\left( {{2^x} + \frac{1}{{{4^x}}}} \right)^n\,$ ratio of 2nd and third terms is given by

$\,{t_3}/{t_2} = 7$ and the sum of the co-efficients of 2nd and 3rd term is

$36,$ then the value of

$x$ is

0%
$\dfrac{-1}{3}$
0%
$\dfrac{-1}{2}$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$

Explanation

Consider the given expression.

$\Rightarrow {{\left( {{2}^{x}}+\dfrac{1}{{{4}^{x}}} \right)}^{n}}$

Given:

$\dfrac{{{t}_{3}}}{{{t}_{2}}}=7$

By binomial expansion, we know that

${{t}_{3}}={}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}$

${{t}_{2}}={}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}$

Again, we have

${}^{n}{{C}_{1}}+{}^{n}{{C}_{2}}=36$

${}^{n+1}{{C}_{2}}=36$

$\dfrac{\left( n+1 \right)!}{2!\left( n+1-2 \right)!}=36$

$\dfrac{n\left( n+1 \right)}{2}=36$

${{n}^{2}}+n-72=0$

$n=8,-9$

Therefore,

$n=8$

Now,

$\dfrac{{{t}_{3}}}{{{t}_{2}}}=\dfrac{{}^{n}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{n-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{n}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{n-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}}$

$\dfrac{{}^{8}{{C}_{2}}{{\left( {{2}^{x}} \right)}^{8-2}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{2}}}{{}^{8}{{C}_{1}}{{\left( {{2}^{x}} \right)}^{8-1}}{{\left( \dfrac{1}{{{4}^{x}}} \right)}^{1}}}=7$

$\dfrac{28{{\left( {{2}^{x}} \right)}^{6}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{2}}}{8{{\left( {{2}^{x}} \right)}^{7}}{{\left( \dfrac{1}{{{\left( {{2}^{2}} \right)}^{x}}} \right)}^{1}}}=7$

$\dfrac{7\times {{2}^{6x-4x}}}{2\times {{2}^{7x-2x}}}=7$

${{2}^{2x-5x}}=2$

$-3x=1$

$x=-\dfrac{1}{3}$

Hence, this is the required result.

The sum of the binomial coefficients in the expansion of

${ \left( { x }^{ -3/4 }+a{ x }^{ 5/4 } \right) }^{ n }$ lies between

$200$ and

$400$ and the term independent of

$x$ equals

$448$ . The value of

$a$ is

0%
$1$
0%
$2$
0%
$1/2$
0%
for no value of $a$

Explanation

$\left(x^{-\dfrac{3}{4}}+ax^{5/4}\right)^n$

$={^{n}C_0}\left(x^{-3/4}\right)^n+{^{n}C_1}(x^{-3/4})^{n-1}(ax^{5/4})^1.......+{^{n}C_n}(ax^{5/4})^n$

Put

$x=1$

$(1+a)^n={^{n}C_0}+{^{n}C_1}a+{^{n}C_2}a^2+......+{^{n}C_n}a^n$

Put

$a=1$

$2^n={^{n}C_0}+{^{n}C_1}+{^{n}C_2}+........+{^{n}C_n}$

$200 < 2^n < 400$

$[\because 2^8=256]$

$n=8$

$(x^{-3/4}+ax^{5/4})^8$

$T_{r+1}={^8C_r}(x^{-3/4})^{8-r}(ax^{5/4})^r$

$\Rightarrow T_{3+1}={^8C_3}a^3=448$

$T_4=^8C_3a^3=448$

$\Rightarrow \dfrac{8\times 7\times 6}{3\times 2}=a^3=448\Rightarrow a^3=8$

$a=2$

$\begin{bmatrix} \because \dfrac{-3}{4}(8-r)+\dfrac{5r}{4}=0\\ -6+\dfrac{3r}{4}+\dfrac{5r}{4}=0\\ -6+2r=0\\ r=3\end{bmatrix}$ .

The coefficient

${x^n}$ in the expression of

${\left( {1 + x} \right)^{2n}}$ and

${\left( {1 + x} \right)^{2n - 1}}$ are in the ratio.

0%
$1:2$
0%
$1:3$
0%
$3:1$
0%
$2:1$

Explanation

We know that,

General term of

$(a+b)^n$ is

$T_r+1=^nC_r a^{n-r},b^r$

For

$(1+x)^{2n}$

General term of

$(1+x)^{2n}$

Put

$a=1,b=x,n=2n$

$T_r+1=^{2n}C_r(1)^{2n-r}.(x)^r$

$TY_r+1=^{2n}C_r.(x)^r$ ......(i)

For coedfficient of

$x^n$

Put

$r=n$ in

$eq^n$ (i)

$T_r+1=^{2n}C_nx^n$

Coefficient of

$x^n=^{2n}C_n$

$\therefore 2^nC_n=\dfrac{2n!}{n!(2n-n)!}$

$=\dfrac{2n!}{n!(n)!}$

For

$(1+x)^{2n-1}$

General term of

$(1+x)^{2n-1}$

Put

$a=1,b=x,n=2n-1$

$T_r+1=^{2r-1}C^r.(1)^{2n-1-r}.x^r$

$T_r+1=^{2n-1}C_rx^r$ .....(ii)

For coefficient of

$x^n$

Put

$r=n$ in equation (ii)

$T_r+1=^{2n-1}C_n\,x^n$

Coefficient of

$x^n$

$=2^{n-1}C_n\times 2$

$=2\times \dfrac{(2n-1)!}{n!(2n-1-n)!}$

$=2\times \dfrac{(2n-1)!}{n!(n-1)!}$

Multiply and divide by

$n$ ,

we get,

$=\dfrac{2n(2n-1)!}{n!n(n-1)!}$

$=\dfrac{(2n)!}{n!n!}$

Hence, ration is

$1:2$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 11 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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