MCQ Questions for Class 11 Engineering Maths Binomial Theorem Quiz 4

Determine the value of

$x$ in the expression

${ \left( x+{ x }^{ t } \right) }^{ 5 }$ , if the third term in the expression is 10,00,000 where

$t=\log _{ 10 }{ x }$ .

0%
$10\quad$
0%
${ 10 }^{ -5/2 }$
0%
Both (A)and (B)
0%
None of these

Explanation

$T_{3}$

$=\:^{5}C_{2}x^{3+2t}$

$=10^{6}$

$x^{3+2t}=10^{5}$

$\ln(x)(3+2t)=5\ln(10)$

$\log_{10}(x)(3+2t)=5$

$t(3+2t)=5$

$2t^{2}+3t-5=0$

$t=1$ and

$t=\dfrac{-5}{2}$
Hence,

$x=10$ and

$x=10^{\frac{-5}{2}}$ .

The number of terms whose values depends on

$x$ in the expansion of

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

0%
$2n+1$
0%
$2n$
0%
$n$
0%
none of these

Explanation

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

$=[(x-\frac{1}{x})^{2}]^{n}$

$=(x-\frac{1}{x})^{2n}$
Hence there will be 2n+1 terms.
The middle term i.e

$n+1 ^{th}$ term will be independent of x.
Hence total number of terms, dependent on x will be

$2n+1-(1)$

$=2n$ terms.

In the expansion of the expression

${ \left( x+a \right) }^{ 15 }$ , if the eleventh term in the geometric mean of the eighth and twelfth terms, which term in the expression is the greatest?

0%
${T}_{6}$
0%
${T}_{7}$
0%
${T}_{8}$
0%
${T}_{9}$

Explanation

From the given condition, one get

$\displaystyle { \left( _{ }^{ 15 }{ { C }_{ 10 }{ x }^{ 2 }{ a }^{ 10 } } \right) }^{ 2 }=\left( _{ }^{ 15 }{ { C }_{ 7 }{ x }^{ 8 }{ a }^{ 7 } } \right) \left( _{ }^{ 15 }{ { C }_{ 11 }{ x }^{ 4 }{ a }^{ 11 } } \right) \Rightarrow \frac { a }{ x } =\sqrt { \frac { 75 }{ 77 } }$
Now

$\displaystyle \left( x+a \right) ^{ 15 }={ x }^{ 15 }{ \left( 1+\frac { a }{ x } \right) }^{ 15 }$
consider the expansion of

$\displaystyle { \left( 1+\frac { a }{ x } \right) }^{ 15 }$

$\displaystyle { T }_{ r }={ T }_{ r+1 }\Rightarrow \frac { r }{ \left( n-r+1 \right) \left( \frac { a }{ x } \right) } =1$

$\displaystyle \Rightarrow r=16\sqrt { \frac { 75 }{ 77 } } -r\sqrt { \frac { 75 }{ 77 } } \Rightarrow r=\frac { 16\sqrt { \frac { 75 }{ 77 } } }{ 1+\sqrt { \frac { 75 }{ 77 } } } =7$
If

$r=7\Rightarrow { T }_{ 7 }<{ T }_{ 8 }$ and

$r=8\Rightarrow { T }_{ 8 }>{ T }_{ 9 }$
Hence

${ T }_{ 8 }$ is the greater term

The value of

$\cfrac { 1 }{ \left( n-1 \right) ! } +\cfrac { 1 }{ \left( n-3 \right) !3! } +\cfrac { 1 }{ \left( n-5 \right) !5! }+....$

0%
$\cfrac { { 2 }^{ n+1 } }{ n! }$
0%
$\cfrac { { 2 }^{ n-1 } }{ n! }$
0%
$\cfrac { { 2 }^{ n-1 } }{ n+1! }$
0%
$\cfrac { { 2 }^{ n+1 } }{ n-1! }$

Explanation

$\cfrac { 1 }{ \left( n-1 \right) ! } +\cfrac { 1 }{ \left( n-3 \right) !3! } +\cfrac { 1 }{ \left( n-5 \right) !5! }+....$

$=\cfrac{1}{n!}\left(\cfrac { n! }{ \left( n-1 \right) ! 1! } +\cfrac { n! }{ \left( n-3 \right) !3! } +\cfrac { n! }{ \left( n-5 \right) !5! }+.... \right)$

$=\cfrac{1}{n!}( ^nC_1+^nC_3+^nC_5+.............)=\cfrac{2^{n-1}}{n!}$

${ \left( 1+x \right) }^{ n }={ C }_{ 0 }+{ C }_{ 1 }x+{ C }_{ 2 }{ x }^{ 2 }+..........+{ C }_{ n }{ x }^{ R }$ , then the sum

${ C }_{ 0 }+({ C }_{ 0 }+{ C }_{ 1 })+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 })+.....+({ C }_{ 0 }+{ C }_{ 1 }+{ C }_{ 2 }+.....+{ C }_{ n-1 }$ is

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

Explanation

Simplifying, we get

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

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

$=n.2^{n}-n2^{n-1}$

$=n2^{n-1}(2-1)$

$=n2^{n-1}$ .

$n$ is a positive integer and

${ C }_{ k }={ _{ }^{ n }{ C } }_{ k }$ , find the value of

$\sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \cfrac { { C }_{ k } }{ { C }_{ k-1 } } \right) }^{ 2 } }$

0%
$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n+2) }{ 12 }$
0%
$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n-2) }{ 12 }$
0%
$\cfrac { n{ \left( n+1 \right) }^{ 2 }(n+2) }{ 12 }$
0%
$\cfrac { n{ \left( n-1 \right) }^{ 2 }(n-2) }{ 12 }$

Explanation

$\displaystyle \sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \cfrac { { C }_{ k } }{ { C }_{ k-1 } } \right) }^{ 2 } } =\sum _{ k=1 }^{ n }{ { k }^{ 3 }{ \left( \frac { n-k+1 }{ k } \right) }^{ 2 } }$

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

$={ \left( n+1 \right) }^{ 2 }\sum _{ k=1 }^{ n }{ { k } } +\sum _{ k=1 }^{ n }{ { k }^{ 3 } } -2\left( n+1 \right) \sum _{ k=1 }^{ n }{ { k }^{ 2 } }$

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

$=\cfrac { n{ \left( n+1 \right) }^{ 2 }(n+2) }{ 12 }$

The sum of coefficients of

$x^{2r},r=1,2,3,...,$ in the expansion of

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

0%
$2^{n}$
0%
$2^{n-1}-1$
0%
$2^{n}-1$
0%
$2^{n-1}+1$

Explanation

$(1+x)^{n}=a_{0}+a_{1}x+a_{2}x^2+...a_{n}x^{n}$
For

$x=1$ we get

$2^{n}=a_{0}+a_{1}+a_{2}+...a_{n}$ ...(i)
Substituting x=-1, we get

$0=a_{0}-a_{1}+a_{2}+...-a_{n-1}+a_{n}$ ...(ii)
Adding i and ii, we get

$2^{n}=2(a_{0}+a_{2}+a_{4}+...a_{2r})$

$2^{n-1}=a_{0}+a_{2}+a_{4}+...a_{2r}$
Now we know that

$a_{0}=\:^{n}C_{0}=1$
By substituting, we get

$a_{2}+a_{4}+...a_{2r}+..=2^{n-1}-1$

Hence the required answer is

$2^{n-1}-1$

The number of terms with integral coefficients in the expansion of

$\left ( 7^{1/3}+5^{1/2}.x \right )^{600}$ is

0%
$100$
0%
$50$
0%
$101$
0%
none of these

Explanation

The number of integral terms will be

$1+\dfrac{100}{L.C.M(2,3)}$

$=1+\dfrac{600}{6}$

$=1+100$

$=101$

The sum of coefficients in the binomial expansion of

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

$6561$ .The constant term in the expansion is

0%
$^{8}\textrm{C}_{4}$
0%
$16\cdot^{8}\textrm{C}_{4}$
0%
$^{6}\textrm{C}_{4}\cdot 2^{4}$
0%
none of these

Explanation

Substituting

$x=1$ to get the sum of the binomial coefficients
we get

$(3)^{n}=6561$
Taking

$log_{3}$ on both the sides, we get

$n=8$
Therefore the above question reduces to

$(x^{-1}+2x)^{8}$
The general term will be

$T_{r+1}=\:^{8}C_{r}x^{2r-8}2^{r}$
The term independent of

$x$ will be given by

$2r-8=0$

$r=4$
Substituting we get

$\:^{8}C_{4}2^{4}$

$=16(\:^{8}C_{4})$

If the

$4th$ term in the expansion is of

$\left (px+x^{-1} \right )^{m}$ is

$2.5$ for all

$x\epsilon R$ then

0%
$p=\displaystyle\frac{5}{2},m=3$
0%
$p=\displaystyle\frac{1}{2},m=6$
0%
$p=\displaystyle-\frac{1}{2},m=6$
0%
none of these

Explanation

The fourth term has a value 2.5 for all Real x.
Hence

$T_{3+1}=\:^{m}C_{3}p^{m-3}x^{m-3}x^{-3}$
It has to be a term independent of x
Therefore

$m-6=0$

$m=6$
Subtituiting we get

$\:^{6}C_{3}p^{3}=2.5$

$20p^{3}=2.5$

$p=0.5$

The sum

$^{10}\textrm{C}_{3}+^{11}\textrm{C}_{3}+^{12}\textrm{C}_{3}+..........+^{20}\textrm{C}_{3}$ is equal to

0%
$^{21}\textrm{C}_{4}$
0%
$^{21}\textrm{C}_{4}+$ $^{10}\textrm{C}_{4}$
0%
$^{21}\textrm{C}_{17}-$ $^{10}\textrm{C}_{6}$
0%
none of these

Explanation

Using

$\:^{n}C_{r}+\:^{n+1}C_{r}=\:^{n+1}C_{r+1}$

$\:^{10}C_{3}+\:^{11}C_{3}+\:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}$

$=[\:^{10}C_{4}+:^{10}C_{3}+\:^{11}C_{3}+\:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$

$=[\:^{11}C_{4}+:^{11}C_{3}+\:^{12}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$

$=[\:^{12}C_{4}+:^{12}C_{3}+\:^{13}C_{3}+...\:^{20}C_{3}]-\:^{10}C_{4}$
:
:
:

$=\:^{20}C_{4}+\:^{20}C_{3}-\:^{10}C_{4}$

$=\:^{21}C_{4}-\:^{10}C_{4}$

$=\:^{21}C_{17}-\:^{10}C_{6}$

The sum of coefficients of all the integral powers of

$x$ in the expansion of

$(1+2\sqrt{x})^{40}$ is

0%
$3^{40}+1$
0%
$3^{40}-1$
0%
$\displaystyle\frac{1}{2}$ $(3^{40}-1)$
0%
$\displaystyle\frac{1}{2}$ $(3^{40}+1)$

Explanation

$(1+2\sqrt{x})^{40}=1+\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}+\:^{40}C_{3}(2\sqrt{x})^{3}+...+\:^{40}C_{40}(2\sqrt{x})^{40}$
Hence, all the odd terms will have integral powers of

$x$ .
Substituting

$x=1$ , we get

$3^{40}=a_{0}+a_{1}+a_{2}+a_{3}...+a_{40}$ ...(i)
Now consider,

$(1-2\sqrt{x})^{40}=1-\:^{40}C_{1}(2\sqrt{x})^{1}+\:^{40}C_{2}(2\sqrt{x})^{2}-\:^{40}C_{3}(2\sqrt{x})^{3}...+\:^{40}C_{40}(2\sqrt{x})^{40}$
Taking

$x$ as

$1$ , we get

$1=a_{0}-a_{1}+a_{2}-a_{3}...+a_{40}$ ...(ii)
Adding i and, ii we get

$3^{40}+1=2[a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}]$

$\dfrac{3^{40}+1}{2}=a_{0}+a_{2}+a_{4}+a_{6}...+a_{40}$
Hence the sum of the coefficients, of integral powers of

$x$ will be

$\dfrac{3^{40}+1}{2}$
Hence, option D is correct

The absolute value of middle term in the expansion of

$\displaystyle \left ( 1-\frac{1}{x} \right )^{n}.(1-x)^{n}$ is

0%
$^{2n}\textrm{C}_{n}$
0%
$-\ ^{2n}\textrm{C}_{n}$
0%
$-\ ^{2n}\textrm{C}_{n-1}$
0%
none of these

Explanation

Given,

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

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

$=\dfrac{(-1)^{n}}{x^{n}}[1-x]^{2n}$
Therefore the middle term will be

$T_{n+1}=\:^{2n}C_{n}x^{n}\left(\dfrac{(-1)^{n}}{x^{n}}\right)$

$=(-1)^{n}\:^{2n}C_{n}$

The sum of last ten coeffficients in the expansion of

$(1+x)^{19}$ when expanded in ascending powers of

$x$ is

0%
$2^{18}$
0%
$2^{19}$
0%
$2^{18}-$ $^{19}\textrm{C}_{10}$
0%
none of these

Explanation

The last ten coefficients are

$\:^{19}C_{19},\:^{19}C_{18}...\:^{19}C_{11},\:^{19}C_{10}$

Hence

$S=\:^{19}C_{19}+\:^{19}C_{18}+...\:^{19}C_{10}$ .

Now we know, that sum of all coefficients is

$2^{19}$ , that is

$\:^{19}C_{0}+\:^{19}C_{1}+...\:^{19}C_{18}+\:^{19}C_{19}=2^{19}$ .

Also,

$\:^{n}C_{r}=\:^{n}C_{n-r}$

Hence

$\:^{19}C_{0}+\:^{19}C_{1}...+\:^{19}C_{9}+\:^{19}C_{10}+...\:^{19}C_{18}+\:^{19}C_{19}=2^{19}$

$2(\:^{19}C_{10}+\:^{19}C_{11}+...\:^{19}C_{18}+\:^{19}C_{19})=2^{19}$

$2S=2^{19}$

$S=2^{18}$ .

The middle term in the expansion of

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

0%
$^{2n}\textrm{C}_{n}$
0%
$(-1)^{n}[(2n!)/(n!)^{2}].x^{-n}$
0%
$^{2n}\textrm{C}_{n}$ . $\displaystyle \frac{1}{x^{n}}$
0%
none of these

Explanation

Middle term will be

$nth$ term.
Hence

$T_{n+1}$

$=(-1)^{n}\:^{2n}C_{n}\dfrac{2}{3}^{n}.\dfrac{3}{2}^{n}.x^{-2n+n}$

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

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

The sum

$^{20}\textrm{C}_{0}+^{20}\textrm{C}_{1}+^{20}\textrm{C}_{2}+^{20}\textrm{C}_{10}$ is equal to

0%
$\displaystyle2^{20}+\frac{20!}{(10!)^{2}}$
0%
$2^{19}-$ $\displaystyle\frac{1}{(2)}\cdot\frac{20!}{(10!)^{2}}$
0%
$\displaystyle2^{19}+$ $^{20}\textrm{C}_{10}$
0%
$none\ of\ these$

Explanation

$\:^{20}C_{0}+\:^{20}C_{1}+\:^{20}C_{2}+\:^{20}C_{10}$

$=1+20+\dfrac{20(19)}{2}+\dfrac{20!}{10!(10!)}$

$=21+190+\dfrac{20!}{(10!)^{2}}$

$=211+\dfrac{20!}{(10!)^{2}}$
Hence the answer is none of these.

The number of non-zero terms in the expansion of

${ \left( 1+3\sqrt { 2 } x \right) }^{ 9 }+{ \left( 1-3\sqrt { 2 } x \right) }^{ 9 }$ is

0%
9
0%
0
0%
5
0%
10

Explanation

In the expansion of

${ \left( 1+3\sqrt { 2 } x \right) }^{ 9 }+{ \left( 1-3\sqrt { 2 } x \right) }^{ 9 }$

$2nd,4th,6th,8th$ and

$10th$ terms get cancelled.

$\therefore$ Number of non-zero terms in

$2\left[ ^{ 9 }{ { C }_{ 0 } }+^{ 9 }{ { C }_{ 2 } }{ \left( 3\sqrt { 2 } x \right) }^{ 2 }+...+^{ 9 }{ { C }_{ 8 } }{ \left( 3\sqrt { 2 } x \right) }^{ 8 } \right]$ is

$5$

The value of

$\displaystyle \sum_{j=1}^{n}(^{n+1}C_{j}-^{n}C_{j})$ is equal to

0%
$\displaystyle 2^{n}$
0%
$\displaystyle 2^{n}+1$
0%
$\displaystyle 3\cdot 2^{n}$
0%
$\displaystyle 2^{n} -1$

Explanation

Simplifying, we get

$(\:^{n+1}C_{1}+\:^{n+1}C_{2}+...\:^{n+1}C_{n})-(\:^{n}C_{1}+\:^{n}C_{2}+...\:^{n}C_{n})$

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

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

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

$=2^{n}-1$

The sum

$\displaystyle\frac{1}{2}$

$^{10}\textrm{C}_{0}-$

$^{10}\textrm{C}_{1}+$

$2\cdot ^{10}\textrm{C}_{2}-$

$2^{2}\cdot ^{10}\textrm{C}_{3}$ +...+

$2^{9}\cdot ^{10}\textrm{C}_{10}$ is equal to

0%
$\displaystyle\frac{1}{2}$
0%
$0$
0%
$\displaystyle\frac{1}{2}\cdot3^{10}$
0%
none of these

Explanation

Let

$S_{n}=\frac{1}{2}\:^{10}C_{0}-\:^{10}C_{1}+...\:^{10}C_{10}2^{9}$

$2S_{n}=\:^{10}C_{0}-2\:^{10}C_{1}+2^{2}\:^{10}C_{2}+...\:^{10}C_{10}2^{10}$

$2S_{n}=(1-2)^{10}$

$2S_{n}=(1)$

$S_{n}=\frac{1}{2}$

The number of integral terms in the expansion of

$\displaystyle \left ( \sqrt{3}+\sqrt[5]{5} \right )^{256}$ is

0%
$25$
0%
$26$
0%
$24$
0%
None of these

Explanation

Total number of integral terms will be

$\cfrac{256}{L.C.M(2,5)}+1$

$=\cfrac{256}{10}+1$

$=25.6+1$
However

$25.6$ is not an integer, hence take

$[25.6]=25$ where

$[x]$ is the greatest integer function.
Hence total number of integral term, will be

$25+1$

$=26$

$(1+x)^{2n}=a_{0}+a_{1}x+a_{2}x^{2}+...+a_{2n}x^{2n}$ then

0%
$a_{n+1}>$ $a_{n}$
0%
$a_{n+1}<$ $a_{n}$
0%
$a_{n-3}=a_{n+3}$
0%
none of these

Explanation

In the above given binomial expansion

$a_{r}=\:^{2n}C_{r}$

Therefore

$a_{n+1}=\:^{2n}C_{n+1}$

And

$a_{n}=\:^{2n}C_{n}$

Since

$\:^{2n}C_{n}$ is greatest coefficient (because it is the coefficient of the middle term), hence obviously

$\:^{2n}C_{n}>\:^{2n}C_{n+1}$

$a_{n}>a_{n+1}$

Also

$a_{n+3}$

$=\:^{2n}C_{n+3}$

$=\:^{2n}C_{2n-(n+3)}$

$=\:^{2n}C_{n-3}$

$=a_{n-3}$ .

Hence

$a_{n+3}=a_{n-3}$ .

0%
Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is True; Statement-2 is Not a correct explanation for Statement-1
0%
Statement-1 is True, Statement-2 is False
0%
Statement-1 is False, Statement-2 is True

Explanation

Let

$x=1$

Therefore

$a_{0}+a_{1}+a_{2}...a_{36}=2^{36}$

Let

$x=w$

Therefore

$a_{0}+a_{1}w+a_{2}w^{2}+a_{3}w^{3}+a_{4}w^{4}....a_{36}w^{36}=(1+w^{2})^{36}=(-w)^{36}=1$

Let

$x=w^{2}$

Therefore

$a_{0}+a_{1}w^{2}+a_{2}w^{4}+a_{3}w^{6}+a_{4}w^{8}....a_{36}w^{72}=(1+w^{4})^{36}=(-w^{2})^{36}=1$

Adding all them and then simplifying gives us

$3a_{0}+a_{1}(1+w+w^{2})+a_{2}(1+w^{2}+w^{4})+3a_{3}+....=2^{36}+2$

$3a_{0}+a_{1}(1+w+w^{2})+a_{2}(1+w+w^{2})+3a_{3}+....=2^{36}+2$

$3(a_{0}+a_{3}+a_{6}+a_{9}+...a_{36})=2(2^{35}+1)$

$a_{0}+a_{3}+a_{6}+a_{9}+...a_{36}=\dfrac{2(2^{35+1})}{3}$

Let

$n\epsilon N$ .If

$(1+x)^{n}=a_{0}x+a_{1}x+a_{2}x^{2}+...+a_{n}x^{n},$ and

$a_{n-3},a_{n-2},a_{n-1},$ are in AP then

0%
$a_{1},a_{2},a_{3}$ are in $AP$
0%
$a_{1},a_{2},a_{3}$ are in $HP$
0%
$n=7$
0%
$n=14$

Explanation

$a_{n-3},a_{n-2},a_{n-1}$ are in A.P indicates that

$a_{3},a_{2},a_{1}$ are in A.P.
Since they are binomial coefficients.
For the above terms to be in A.P, they must follow the relationship

$(n-2r)^{2}=n+2$
Here r will be 2.
Therefore

$n^2-8r+16=n+2$

$n^2-9r+14=0$

$(n-7)(n-2)=0$

$n=7$ since

$n=2$ is not possible here.

In the expansion of

$\left (\sqrt[3]{4}+\dfrac{1}{\sqrt[4]{6}}\right )^{20},$

0%
the number of rational terms $=4$
0%
the number of irrational terms $=19$
0%
the middle term is irrational
0%
the number of irrational terms $=17$

Explanation

$T_{r+1}=\:^{n}C_r4^{\tfrac{20-r}{3}}6^{\tfrac{-r}{4}}$

$=\:^{n}C_{r}2^{\tfrac{40-2r}{3}-\tfrac{r}{4}}3^{-\tfrac{r}{4}}$

$=\:^{n}C_{r}2^{\tfrac{160-11r}{12}}3^{-\tfrac{r}{4}}$

There are total of

$21$ terms.

Hence, we get rational terms for

$r=20,8$

Hence there are in total

$21-2$

$=19$ irrational terms.

The middle term is at

$r=10$ which is irrational.

The sum of the series

$\sum _{ r=0 }^{ 10 }{ _{ }^{ 20 }{ { C }_{ r }^{ } } }$ is

0%
$\displaystyle { 2 }^{ 19 }-\frac { 1 }{ 2 } ._{ }^{ 20 }{ { C }_{ 10 }^{ } }$
0%
$\displaystyle { 2 }^{ 19 }+\frac { 1 }{ 2 } .{ }^{ 20 }{ { C }_{ 10 }^{ } }$
0%
${ 2 }^{ 19 }$
0%
${ 2 }^{ 20 }$

Explanation

We have,

$\sum _{ r=0 }^{ 10 }{ ^{ 20 }{ { C }_{ r } } } =^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 10 } }$

But

$^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 20 } }={ 2 }^{ 20 }$

and

$\because \quad ^{ 20 }{ { C }_{ 20 } }=^{ 20 }{ { C }_{ 0 } };^{ 20 }{ { C }_{ 19 } }=^{ 20 }{ { C }_{ 1 } };^{ 20 }{ { C }_{ 18 } }=^{ 20 }{ { C }_{ 2 } }...$ and

$^{ 20 }{ { C }_{ 11 } }=^{ 20 }{ { C }_{ 9 } }$

$\displaystyle \therefore \sum _{ r=0 }^{ 10 }{ ^{ 20 }{ { C }_{ r } }= } \left( ^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 20 } } \right) -\left( ^{ 20 }{ { C }_{ 11 } }+^{ 20 }{ { C }_{ 12 } }+...+^{ 20 }{ { C }_{ 20 } } \right) \\ ={ 2 }^{ 20 }+^{ 20 }{ { C }_{ 10 } }-\left( ^{ 20 }{ { C }_{ 10 } }+^{ 20 }{ { C }_{ 9 } }+...+^{ 20 }{ { C }_{ 0 } } \right) \\ \Rightarrow 2\left[ ^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 10 } } \right] ={ 2 }^{ 20 }+^{ 20 }{ { C }_{ 10 } }$

$\displaystyle \therefore ^{ 20 }{ { C }_{ 0 } }+^{ 20 }{ { C }_{ 1 } }+...+^{ 20 }{ { C }_{ 10 } }={ 2 }^{ 19 }+\frac { 1 }{ 2 } .^{ 20 }{ { C }_{ 10 } }$

$^{ n+1 }{ { C }_{ 2 }^{ } }+2\left[ _{ }^{ 2 }{ { C }_{ 2 }^{ } }+^{ 3 }{ { C }_{ 2 }^{ } }+^{ 4 }{ { C }_{ 2 }^{ } }+...+^{ n }{ { C }_{ 2 }^{ } } \right] =$

0%
$\displaystyle \frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$
0%
$\displaystyle \frac { n\left( n+1 \right) }{ 2 }$
0%
$\displaystyle \frac { n\left( n-1 \right) \left( 2n-1 \right) }{ 6 }$
0%
None of these

Explanation

We have,

$^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 2 }{ { C }_{ 2 } }+^{ 3 }{ { C }_{ 2 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 3 }{ { C }_{ 3 } }+^{ 3 }{ { C }_{ 2 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] \\ =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 4 }{ { C }_{ 3 } }+^{ 4 }{ { C }_{ 2 } }+...+^{ n }{ { C }_{ 2 } } \right] =^{ n+1 }{ { C }_{ 2 } }+2\left[ ^{ 5 }{ { C }_{ 3 } }+...+^{ n }{ { C }_{ 2 } } \right] \\ =^{ n+1 }{ { C }_{ 2 } }+2.^{ n+1 }{ { C }_{ 3 } }=^{ n+1 }{ { C }_{ 2 } }+^{ n+1 }{ { C }_{ 3 } }+^{ n+1 }{ { C }_{ 3 } }=^{ n+2 }{ { C }_{ 3 } }+^{ n+1 }{ { C }_{ 3 } }$

$\displaystyle =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 6 } +\frac { n\left( n+1 \right) \left( n-1 \right) }{ 6 } =\frac { n\left( n+1 \right) \left( 2n+1 \right) }{ 6 }$

$\displaystyle C_{0},C_{1},C_{2},....C_{n}$ are binomial coefficient in the expansion of

$\displaystyle (1+x)^{n}$ , then value of

$\displaystyle C_{1}+C_{4}+C_{7}+...$ equals

0%
$\displaystyle \frac{1}{3}(2^{n}+\sqrt 3 \sin \frac{n \pi}{3})$
0%
$\displaystyle \frac{1}{3}(2^{n}-\cos \frac{n \pi}{3}+\sqrt 3 \sin \frac{n \pi}{3} )$
0%
$\displaystyle \frac{1}{3}(2^{n}-\sqrt 3 \sin \frac{n \pi}{3})$
0%
$\displaystyle \frac{1}{3}(2^{n}-\cos \frac{n \pi}{3}-\sqrt 3 \sin \frac{n \pi}{3} )$

Explanation

Given $\displaystyle \left ( 1+x \right )^{n}=C_{0}+C_{1}x+C_{2}x^{2}+\cdots +C_{n}x^{n}$
$\displaystyle \therefore x^{2}\left ( 1+x \right )^{n}=x^{2}C_{0}+C_{1}x^{3}+C_{2}x^{4}+\cdots +C_{n}x^{n+2}\left ( * \right )$
Now substituting $x=1$ , $\displaystyle \omega , \omega ^{2}in\left ( * \right ),$ we get $\displaystyle 2^{n}=C_{0}+C_{1}+C_{2}+C_{3}+\cdots +C_{n}\cdots \left ( 1 \right )$
$\displaystyle \omega ^{2}\left ( 1+\omega \right )^{n}=C_{0}\omega ^{2}+C_{1}\omega ^{3}+C_{2}\omega ^{4}+\cdots +C_{n}.\omega ^{n+2}$ ...(2)
$\displaystyle \omega ^{4}\left ( 1+\omega ^{2} \right )^{n}=C_{0}\omega ^{4}+C_{1}\omega ^{6}+C_{2}\omega ^{8}+\cdots +C_{n}.\omega ^{2n+4}$ ...(3)
Now adding (1), (2) and (3) we get $\displaystyle 2^{n}+\omega, ^{2}\left ( 1+\omega \right )^{n}+\omega ^{4}\left ( 1+\omega ^{2} \right )^{n}$
$\displaystyle =C_{0}\left ( 1+\omega +\omega ^{2} \right )+C_{1}\left ( 1+\omega ^{3}+\omega ^{6} \right )+C_{2}\left ( 1+\omega +\omega ^{2} \right )$
$\displaystyle +....=3\left ( C_{1}+C_{4}+C_{7}+\cdots \right )$ (other terms vanished). $\displaystyle 3\left ( C_{1}+C_{4}+C_{7}+\cdots \right )$ $\displaystyle =2^{n}+\omega ^{2}\left ( \cos\frac{n\pi }{2}+i \sin\frac{n\pi }{3} \right )+\omega ^{4}\left ( \cos\frac{n\pi }{3}-i \sin \frac{n\pi }{3} \right )$ $\displaystyle =2^{n}+\left ( \omega ^{2}+\omega \right )\cos\frac{n\pi }{3}+i\left ( \omega ^{2}-\omega \right )\sin \frac{n\pi }{3}$ $\displaystyle =2^{n}-\cos\frac{n\pi }{3}+i\left ( -\sqrt{3i} \right )\sin\frac{n\pi }{3}$ As $\displaystyle \omega =\frac{-1+i\sqrt{3}}{2}and \omega ^{2}=\frac{-1-i\sqrt{3}}{2}$ $\displaystyle \omega ^{2}-\omega =-\frac{1}{2}-\frac{i\sqrt{3}}{2}+\frac{1}{2}-\frac{i\sqrt{3}}{2}=i\sqrt{3}$ $\displaystyle \omega ^{2}-\omega =-\sqrt{3}$ $\displaystyle \therefore 3\left ( C_{1}+C_{4}+C_{7}\cdots \right )=2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3}$ $\displaystyle C_{1}+C_{4}+C_{7}+\cdots =\frac{1}{3}\left ( 2^{n}-\cos\frac{n\pi }{3}+\sqrt{3}\sin \frac{n\pi }{3} \right )$

If the third term in the expansion of

$\displaystyle (\frac{1}{x}+x^{\log_{10}x})^{5}$ is

$1,000$ , then

$x$ -equals

0%
10 $\displaystyle ^{2}$
0%
10 $\displaystyle ^{3}$
0%
$10$
0%
None of these

Explanation

$T_{2+1}$

$=T_{3}$

$=\:^{5}C_{2}x^{-3}x^{2log_{10}(x)}$

$=10^{3}$
Therefore

$x^{-3}x^{2log_{10}(x)}=100$

$x^{2log_{10}(x)-3}=100$
Now x has to be in powers of 10.
Therefore
Let

$x=10^k$
Above equation reduces to

$10^{k(2k-3)}=10^{2}$

$2k^{2}-3k-2=0$

$(2k+1)(k-2)=0$

$k=2$ and

$k=\frac{-1}{2}$
Hence

$x=10^{2}$ and

$x=10^{-\frac{1}{2}}$

$C_{0},C_{1},C_{2},.....C_{n},$ are binomial coefficients,then

$\displaystyle\sum_{k= 0}^{n} C_{k}\:\sin kx \cos \left ( n-k \right )x$ equals

0%
$2^{n} \sin nx$
0%
$2^{n+1}\sin \left ( n+1 \right )x$
0%
$2^{n-1}\sin nx$
0%
$2^{n+1}\sin nx$

Explanation

$S=\sum \:^{n}C_{k} sin(kx) cos(n-k)x$
Now writing the terms in reverse, we get

$S=\sum \:^{n}C_{n-k} sin(n-k)x cos(k)x$
Hence

$2S=\sum \:^{n}C_{k} [sin(kx) cos((n-k)x)+cos(kx)sin((n-k)x)]$

$=\sum \:^{n}C_{k} sin(nx)$
Hence

$2S=2^{n} sin(nx)$

$S=2^{n-1}sin(nx)$

If coefficient of

$x^{100}$ in

$1+\left ( 1+x \right )\left ( 1+x \right )^{2}+.....+\left ( 1+x \right )^{n}\left ( if\:n \geq 100\right )$ is

$C_{101}^{201}$ then the value of n equals

0%
$202$
0%
$100$
0%
$200$
0%
$201$

Explanation

${ ^{ n }{ C } }_{ r }+{ ^{ n }{ C } }_{ (r+1) }={ ^{ (n+1) }{ C } }_{ (r+1) }$
coefficient of

${ x }^{ 100 }$ is

$^{ 100 }{ { C }_{ 100 } }+^{ 101 }{ { C }_{ 100 } }+^{ 102 }{ { C }_{ 100 } }+........+^{ n }{ { C }_{ 100 } }$ .
Which is equal to

${ ^{ (n+1) }{ C } }_{ 101 }$ .
Therefore,

$n+1=201$
Which implies

$n=200$

$\displaystyle \left ( 1+x+x^{2} \right )^{n}=\sum_{r=0}^{2n} a_{r}x^{r}=a_{0}+a_{1}x+a_{2}x^{2}+...+a^{2n}x^{2n}$ and

$\displaystyle P=a_{0}+a_{3}+a_{6}+...$

$\displaystyle Q=a_{1}+a_{4}+a_{7}+...$

$\displaystyle R=a_{2}+a_{5}+a_{8}+...$
then the set of values of

$P, Q, R$ are respectively equals

0%
$\displaystyle (1 ,1, 1)$
0%
$\displaystyle (3^{n},3^{n},3^{n})$
0%
$\displaystyle (3^{n+1},3^{n+1},3^{n+1})$
0%
$\displaystyle (3^{n-1},3^{n-1},3^{n-1})$

Explanation

Sum of all coefficients will be

$3^{n}$

$=3(3^{n-1})$
Now
The sum of coefficients of

$a_{0}+a_{3}+a_{6}....$
will be equal to

$a_{1}+a_{4}+a_{7}+....$ which will be further equal to

$a_{2}+a_{5}+a_{8}+...$

$=$ sum of all coefficients/3

$=\dfrac{3(3^{n-1})}{3}=3^{n-1}$

The evaluated value of

$\displaystyle \sum_{i=0}^{n} \sum_{j=1}^{n}$

$\displaystyle ^{n}C_{j}$

$\displaystyle ^{j}C_{i},$

$\displaystyle i\leq j$

0%
$\displaystyle 3^{n}+1$
0%
$\displaystyle 3^{n}-1$
0%
$\displaystyle 3^{n+1}+1$
0%
None of these

Explanation

Expanding, we get

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

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

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

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

$=3^{n}-1$

$\displaystyle C_{r}=^{n}C_{r}$ and

$\displaystyle (C_{0}+C_{1})(C_{1}+C_{2})...(C_{n-1}+C_{n})=k$

$\displaystyle (C_{0} C_{1}C_{2}...C_{n})$ then the value of

$\displaystyle k$ equals

0%
$\displaystyle \frac{(n+1)^{n+1}}{n!}$
0%
$\displaystyle \frac{(n+1)^{n}}{n.n!}$
0%
$\displaystyle \frac{(n)^{n}}{n!}$
0%
$\displaystyle \frac{(n+1)^{n}}{n!}$

Explanation

$\:^{n}C_{r+1}+\:^{n}C_{r}$
$=\:^{n+1}C_{r+1}$
Hence simplifying the terms, we get
$\:^{n+1}C_{1}.\:^{n+1}C_{2}....\:^{n+1}C_{n}$
Now
$\:^{n+1}C_{1}$
$=\dfrac {(n+1)!}{n!.1!}$
$=\dfrac {(n+1)}{n}\:^{n}C_{1}$
Similarly
$\:^{n+1}C_{2}$
$=\dfrac {(n+1)!}{(n-1)!.2!}$
$=\dfrac {(n+1)}{(n-1)}\:^{n}C_{2}$
Hence substituting, in the above expression, we get
$\dfrac {(n+1)^{n}}{n!}(\:^{n}C_{0}.\:^{n}C_{1}...\:^{n}C_{n})$
Comparing coefficients, we get
$K=\dfrac {(n+1)^{n}}{n!}$

$\displaystyle P$ be the sum of odd term and

$\displaystyle Q$ that of even terms in the expansion of

$\displaystyle (x+a)^{n}$ , then the value of

$\displaystyle [(x+a)^{2n}-(x-a)^{2n}]$ equals

0%
$\displaystyle PQ$
0%
$\displaystyle 2PQ$
0%
$4$ $\displaystyle PQ$
0%
None of these

Explanation

$P=\dfrac{(x+a)^{n}+(x-a)^{n}}{2}$ ...(i)

$Q=\dfrac{(x+a)^{n}-(x-a)^{n}}{2}$ ...(ii)
Hence,

$2P.2Q$

$=4PQ$

$=[(x+a)^{n}+(x-a)^{n}][(x+a)^{n}-(x-a)^{n}]$

$=(x+a)^{2n}-(x-a)^{2n}$

The sum of the coefficients of all odd exponets of

$\displaystyle x$ in the product of

$\displaystyle (1-x +x^{2}-x^{3}+x^{4}+...-x^{49}+x^{50})\times(1+x+x^{2}+x^{3}+...+x^{50})$ equals

0%
$1$
0%
$0$
0%
$-1$
0%
None of these

Explanation

Coefficient of

$x^{1}$ will be

$1+(-1)$

$=0$ ..(i)
Coefficient of

$x^{3}$ will be

$1+(-1)+1+(-1)$

$=0$
Hence the coefficients of

$x^{n}$ where n is odd is zero.

$C_{0},C_{1},C_{2}....,C_{n}$ are Binomial Coefficients, such that

$\displaystyle S_{n}=\sum_{r=0}^{n}\frac{1}{{C_{r}}^{n}}$ and

$\displaystyle t_{n}= \sum_{r= 0}^{n}\frac{r}{{C_{r}}^{n}}$ then

$\displaystyle \frac{t_{n}}{s_{n}}$ equals

0%
$\displaystyle \frac{n}{2}$
0%
$\displaystyle\frac{n\left ( n+1 \right )}{2}$
0%
$\displaystyle\frac{n+1}{2}$
0%
None of these

Explanation

$\displaystyle S_{ n }=\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n }{ 2+1 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } +\frac { 1 }{ ^{ n }{ { C }_{ n-r } } } }$

$\displaystyle \Rightarrow { S }_{ n }=\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 2 }{ ^{ n }{ C_{ r } } } =2 } \sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { 1 }{ ^{ n }{ C }_{ r } } }$

$\displaystyle \therefore { t }_{ n }=\sum _{ r=0 }^{ n }{ \frac { r }{ ^{ n }{ C }_{ r } } } =\sum _{ r=0 }^{ \frac { n+1 }{ 2 } }{ \frac { r+\left( n-r \right) }{ ^{ n }{ { C }_{ r } } } }$

$\displaystyle \Rightarrow n\sum _{ r=0 }^{ n }{ \frac { 1 }{ ^{ n }{ C }_{ r } } } =n\left( \frac { { S }_{ n } }{ 2 } \right) =\frac { n }{ 2 } { S }_{ n }$

Thus

$\dfrac{t_n}{S_n} = \dfrac{n}{2}$

$\displaystyle\left ( 1+x \right )^{n}= \sum_{r= 0}^{n}C_{r}x^{r}$ , then the value of

$C_{0}-C_{2}+C_{4}-C_{6}+C_{8}-C_{10}+...$ equals

0%
$\displaystyle2^{\tfrac{n}{2}}\cos \frac{nx}{4}$
0%
$C_{1}-C_{3}+C_{5}-C_{7}+....$
0%
$C_{0}+C_{4}+C_{8}+C_{12}+....$
0%
$\displaystyle2^{\tfrac{n}{2}}\sin \frac{nx}{4}$

Explanation

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

$(1-x)^{n}=\:^{n}C_{0}-\:^{n}C_{1}x+\:^{n}C_{2}x^{2}+...(-1)^{n}\:^{n}C_{n}x^{n}$
Adding, we get

$(1+x)^{n}+(1-x)^{n}=2[\:^{n}C_{0}+\:^{n}C_{2}x^{2}+\:^{n}C_{4}x^{4}+\:^{n}C_{6}x^{6}+...]$
Substituting,

$x=\sqrt{-1}=i$ we get

$(1+i)^{n}+(1-i)^{n}=2[\:^{n}C_{0}-\:^{n}C_{2}+\:^{n}C_{4}x^{4}-\:^{n}C_{6}x^{6}+...]$

$2^{\dfrac{n}{2}-1}[e^{\dfrac{in\pi}{4}}+e^{\dfrac{-in\pi}{4}}]=\:^{n}C_{0}-\:^{n}C_{2}+\:^{n}C_{4}x^{4}-\:^{n}C_{6}x^{6}+...$
Hence
Solving the LHS, we get

$2^{\frac{n}{2}}\cos(\displaystyle \dfrac{n\pi}{4})$

$\left ( 1+x \right )^{n}= \sum_{r= 0}^{n}C_{r}x^{r}$ then the value of

$3C_{1}+7C_{2}+11C^{3}+....+\left ( 4n-1 \right )C_{n}$ is

0%
$\left ( 4n-1 \right )2^{n}$
0%
$\left ( 2n-1 \right )2^{n}$
0%
$\left ( 2n-1 \right )2^{n}+1$
0%
$\left ( 4n-1 \right )2^{n}-1$

Explanation

Let

$n=1$ , we get

$S_{1}=3\:^{1}C_{1}$

$=3$

$=(2(1)-1)2^{1}+1$
Let

$n=2$ , we get

$S_{2}=3\:^{2}C_{1}+7\:^{2}C_{2}$

$=6+7$

$=13$

$=(2(2)-1)2^{2}+1$
Let

$n=3$ , we get

$S_{3}=3\:^{3}C_{1}+7\:^{3}C_{2}+11\:^{3}C_{3}$

$=9+21+11$

$=41$

$=(2(3)-1)2^{3}+1$
Hence

$S_{n}$

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

${C_{r}}^{13}$ denoted by

$C_{r}$ then value of

$c_{1}+c_{5}+c_{7}+c_{9}+c_{11}$ is equal to

0%
$2^{12}-287$
0%
$2^{12}-165$
0%
$2^{12}-C_{3}$
0%
$2^{12}-C_{2}-C_{13}$

Explanation

Consider

$(1+x)^{13}=\:^{13}C_{0}+\:^{13}C_{1}.x^{1}+\:^{13}C_{2}.x^{2}+\:^{13}C_{3}.x^{3}...+\:^{13}C_{13}.x^{13}$
Now Let x=1.

$2^{13}=\:^{13}C_{0}+\:^{13}C_{1}+\:^{13}C_{2}+\:^{13}C_{3}...+\:^{13}C_{13}$

$=2[\:^{13}C_{0}+\:^{13}C_{1}+\:^{13}C_{2}+\:^{13}C_{3}...+\:^{13}C_{6}]$
Hence

$2^{12}=a_{0}+a_{1}+...a_{6}$ ...(i)
Similarly

$(1-x)^{13}|_{x=-1}=a_{0}-a_{1}+a_{2}-a_{3}...a_{6}=0...(ii)$
Adding i and ii, we get

$2^{12}=2(a_{0}+a_{2}+a_{4}+a_{6})$

$=2(a_{0}+a_{11}+a_{9}+a_{7})$
Hence

$2^{11}=a_{0}+a_{11}+a_{9}+a_{7}$ ...(a)
Subtracting ii from i, we get

$2^{12}=2(a_{1}+a_{3}+a_{5})$
Or

$2^{11}=a_{1}+a_{3}+a_{5}$ ...(b)
Adding a and b, we get

$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(a_{0}+a_{3})$

$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(1+\:^{13}C_{3})$

$2^{12}=a_{1}+a_{5}+a_{7}+a_{9}+(287)$

$2^{12}-287=a_{1}+a_{5}+a_{7}+a_{9}$ .

The Coefficient of

$x^{53}$ in

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

0%
$=-\:^{100}C_{53}$
0%
$=-\:^{101}C_{53}$
0%
$=\:^{101}C_{47}$
0%
$=-\:^{100}C_{47}$

Explanation

The general term is

$G$

$=\:^{100}C_{m}(x-3)^{100-m}2^{m}$
This is the general term of

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

$=(x-1)^{100}$

$=(1-x)^{100}$
Hence the coefficient of

$x^{53}$ is

$T_{54}$

$=(-1)^{53}\:^{100}C_{53}$

$=-\:^{100}C_{53}$

The number of terms with integral coefficient in the expansion of

$\left ( 17^{\dfrac 13} +35^{\dfrac 12}\right )^{300}$ is

0%
$50$
0%
$100$
0%
$150$
0%
$51$

Explanation

The number of rational terms will be

$1+\dfrac{300}{L.C.M(3,2)}$

$=1+\dfrac{300}{6}$

$=1+50=51$ rational terms.

$A$ is the sum of the odd terms and

$B$ the sum of even terms in the expansion of

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

${ A }^{ 2 }-{ B }^{ 2 }=$

0%
${ \left( { x }^{ 2 }+{ a }^{ 2 } \right) }^{ n }$
0%
${ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$
0%
$2{ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$
0%
None of these

Explanation

We have,

${ \left( x+a \right) }^{ n }=^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 2 }^{ } }{ x }^{ n-2 }{ a }^{ 2 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+...+^{ n }{ { C }_{ n }^{ } }{ a }^{ n }\\ =\left( ^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 2 }^{ }{ x }^{ n-2 }{ a }^{ 2 } }+... \right) +\left( ^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+... \right) \\ =A+B\\ { \left( x-a \right) }^{ n }=^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }-^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 2 }^{ } }{ x }^{ n-2 }{ a }^{ 2 }-^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }\\ =\left( ^{ n }{ { C }_{ 0 }^{ } }{ x }^{ n }+^{ n }{ { C }_{ 2 }^{ }{ x }^{ n-2 }{ a }^{ 2 } }+... \right) -\left( ^{ n }{ { C }_{ 1 }^{ } }{ x }^{ n-1 }{ a }^{ 1 }+^{ n }{ { C }_{ 3 }^{ } }{ x }^{ n-3 }{ a }^{ 3 }+... \right) \\ =A-B\\ \therefore { A }^{ 2 }-{ B }^{ 2 }=\left( A+B \right) \left( A-B \right) ={ \left( x+a \right) }^{ n }{ \left( x-a \right) }^{ n }={ \left( { x }^{ 2 }-{ a }^{ 2 } \right) }^{ n }$

The sum of the series

$\displaystyle\sum_{r=0}^{n}\left ( ^{n+1}C_{r} \right )$ equals

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

Explanation

Let

$\sum_{r=0}^{n} (^{n+1}C_r)$ be

$f$ .

$f=\sum_{r=0}^n (^{n+1}C_r)$

Consider the expansion of

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

$(1+x)^{n+1}=\sum_{k=0}^{n+1} (^{n+1}C_r 1^{n+1-k} x^{k})$

Substituting

$x=1$ in above equation

$(1+1)^{n+1}=\sum_{k=0}^{n+1} (^{n+1}C_k) = f + ^{n+1}C_{n+1}$

$\therefore f=2^{n+1}-1$

The value of asked expression is

$2^{n+1}-1$ .

$x+y= 1$ then

$\displaystyle\sum_{r= 0}^{n} r^{n}C_{r}x^{r}y^{n-r}$ equals

0%
$1$
0%
$n$
0%
$nx$
0%
$ny$

Explanation

Expanding, we get

$\:^{n}C_{1}xy^{n-1}+2\:^{n}C_{2}x^{2}y^{n-2}+3\:^{n}C_{3}x^{3}y^{n-3}+...n\:^{n}C_{n}x^{n}$

$=x[\:^{n}C_{1}y^{n-1}+2\:^{n}C_{2}x^{1}y^{n-2}+3\:^{n}C_{3}x^{2}y^{n-3}+...n\:^{n}C_{n}x^{n-1}]$

$=x[\dfrac{d(y+x)^{n}}{dx}]$ ...(keeping

$y$ constant.)

$=x(n(x+y)^{n-1})$
Now

$x+y=1$ ,
Hence we get

$x(n)$

$=nx$

Number of rational term is the expansion of

$\left ( 7^{1/3}+11^{1/9} \right )^{729}$

0%
81
0%
82
0%
730
0%
None of these

Explanation

Since 7 and 11 are prime numbers, hence application of general formula for number of rational terms will be

$=1+\dfrac{729}{L.C.M(3,9)}$

$=1+\dfrac{729}{9}$

$=1+81$

$=82$ rational terms.

The values of

$x$ in the expansion

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

$10,00,000$

0%
$\displaystyle 10$
0%
$\displaystyle 10^{2}$
0%
$\displaystyle 10^{3}$
0%
None of these

Explanation

$T_{3}$

$=\:^{5}C_{2}x^{3+2log_{10}(x)}$

$=10^6$

$x^{3+2log_{10}(x)}=10^{5}$
Now x has to be in powers of

$10$ .
Let

$x=10^{k}$
Therefore,

$10^{k(3+2k)}=10^{6}$

$2k^{2}+3k=5$

$2k^{2}+3k-5=0$

$2k^{2}+5k-2k-5=0$

$(k-1)(2k+5)=0$

$k=1$ and

$k=\frac{-5}{2}$
Therefore,

$x=10$ and

$x=10^{\frac{-5}{2}}$

$\displaystyle\left ( 1+x \right )^{n}=\sum_{r=0}^{n}C_{r}x^{r}$ and

$\sum { \sum _{ 0\le i<j\le n }{ { C }_{ i }\times { C }_{ j } } }$ represent the products of the

$C_{i}$ 's taken two at a time, then its value equals

0%
$\displaystyle2^{2n-1}-\frac{\left (2n\right)!}{\left (n!\right)^{2}}$
0%
$\displaystyle2^{2n-1}+\frac{2n!}{n!n!}$
0%
$\displaystyle2^{2n-1}-\frac{2n!}{2\cdot n!n!}$
0%
None of these

Explanation

$A=\sum { \sum _{ 0\le i<j\le 1 }{ { C }_{ i }\times { C }_{ j } } }$
so,

$A=\cfrac { 1 }{ 2 } \times ({ (\sum _{ r=0 }^{ n }{ { C }_{ i }) } }^{ 2 })-\sum _{ r=0 }^{ n }{ { { C }_{ i } }^{ 2 } } )$
therefore our required answer is option C.

$\sum _{ i=0 }^{ n }{ { ({ C }_{ i }) }^{ 2 }= } ^{ 2n }{ { C }_{ n } }$

${ (\sum _{ i=0 }^{ n }{ { C }_{ i } } ) }^{ 2 }={ 2 }^{ 2n }$

Sum of the coefficients of the terms of degree

$m$ in the expansion of

${ (1+x) }^{ n }{ (1+y) }^{ n }{ (1+z) }^{ n }$ is

0%
${ ({ _{ }^{ n }{ C } }_{ m }) }^{ 3 }$
0%
$3({ _{ }^{ n }{ C } }_{ m })$
0%
${ _{ }^{ n }{ C } }_{ 3m }$
0%
${ _{ }^{ 3n }{ C } }_{ m }$

Explanation

The Coefficient of

$x^{m}$

$=$ Number of ways of choosing

$m$ balls out of

$n$ black balls,

$n$ green balls and

$n$ blue ball.
Hence total number of balls

$=3n$ .
Required is

$m$ .
Hence required combination is

$\:^{3n}C_{m}$ .
Hence the coefficient of

$x^{m}$ in

$(1+x)^{n}(1+y)^{n}(1+z)^{n}$

$=\:^{3n}C_{m}$ .

The number of irrational terms in the expansion of

${ ({ 5 }^{\tfrac 16 }+{ 2 }^{ \tfrac 18 }) }^{ 100 }$ is

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

Explanation

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

${ ({ 5 }^{\tfrac 16 }+{ 2 }^{\tfrac 18 }) }^{ 100 }$ is

${ t }_{ r+1 }={ _{ }^{ 100 }{ C } }_{ r }{ ({ 5 }^{\tfrac 16 }) }^{ 100-r }{ ({ 2 }^{\tfrac 18 }) }^{ r }$
As

$5$ and

$2$ are relatively prime,

${t}_{r+1}$ will be rational if

$\cfrac { 100-r }{ 6 }$ and

$\cfrac { r }{ 8 }$ are both integers. i.e., if

$100-r$ is a multiple of

$6$ and

$r$ is a multiple of

$8$ . As

$0\le r \le 100$ , multiples of

$8$ upto

$100$ and corresponding value of

$100-r$ are

$r=0,8,16,24,.....88,96$

$100-r=100,92,84,76,....12,4$
Out of values of

$100-r$ , multiples of

$6$ are

$84,60,36,12$

$\therefore$ there are just four rational terms.

$\Rightarrow$ number of irrational terms is

$101-4=97$

Value of the expression

${ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+.....+(n+1){ C }_{ n }^{ 2 }$ is

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

Explanation

Let

$S = { C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+{ C }_{ 2 }^{ 2 }+.....n{ C }_{ n-1 }^{ 2 }+(n+1){ C }_{ n }^{ 2 }\quad ......(1)$
using

${ C }_{ r }={ C }_{ n-r }$ , we write

$(1)$ in the reverse order to obtain

$S=(n+1){ C }_{ 0 }^{ 2 }+{ C }_{ 1 }^{ 2 }+....2{ C }_{ n-1 }^{ 2 }+{ C }_{ n }^{ 2 }\quad ......(2)$
Adding

$(1)$ and

$(2)$ , we get

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

$\Rightarrow \quad S=\left( \cfrac { n }{ 2 } +1 \right) ({ _{ }^{ 2n }{ C } }_{ n })$

CBSE Questions for Class 11 Engineering Maths Binomial Theorem Quiz 4 - 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 4 - MCQExams.com

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Practice Class 11 Engineering Maths Quiz Questions and Answers

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