ISI2018-MMA-26

Question

Let $C_i(i=0,1,2...n)$ be the coefficient of $x^i$ in $(1+x)^n$.Then 

$\frac{C_0}{2} – \frac{C_1}{3}+\frac{C_2}{4}-\dots +(-1)^n \frac{C_n}{n+2}$ is equal to

• A. $\frac{1}{n+1}\\$
• B. $\frac{1}{n+2}\\$
• C. $\frac{1}{n(n+1)}\\$
• D. $\frac{1}{(n+1)(n+2)}$

Comments

Is 0 is not in options??

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If we take n=2,

$(1+x)^2=1+2x+x^2$

==> $\frac{1}{2}-\frac{2}{3}+\frac{1}{4}=\frac{1}{12}$

Matches with option D.

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yes, but by formula not coming. chk once.

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@Verma Ashish good hack

Answer 1

$\left ( 1+x \right )^{n}=^{n}\textrm{C}_{0}+^{n}\textrm{C}_{1}.x+^{n}\textrm{C}_{2}.x^{2}+^{n}\textrm{C}_{3}.x^{3}+.....+^{n}\textrm{C}_{n}.x^{n}$

Multiplying both sides by $x$

$x\left ( 1+x \right )^{n}=^{n}\textrm{C}_{0}.x+^{n}\textrm{C}_{1}.x^{2}+^{n}\textrm{C}_{2}.x^{3}+^{n}\textrm{C}_{3}.x^{4}+.....+^{n}\textrm{C}_{n}.x^{n+1}$

Integrating both sides w.r.t. $x$ we get

$\frac{x.(1+x)^{(n+1)}}{n+1}-\frac{(1+x)^{(n+2)}}{(n+1).(n+2)}+k=\frac{^{n}\textrm{C}_{0}.x^{2}}{2}+\frac{^{n}\textrm{C}_{1}.x^{3}}{3}+\frac{^{n}\textrm{C}_{2}.x^{4}}{4}+.....+\frac{^{n}\textrm{C}_{n}.x^{n+2}}{n+2}$ , where $k$ is arbitrary constant.

Now putting $x=0$ on both sides, $k=\frac{1}{(n+1)(n+2)}$

Again putting $x=-1$ we get,

$\frac{1}{(n+1)(n+2)}=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-......+(-1)^{n+2}\frac{C_{n}}{n+2}$

$\frac{1}{(n+1)(n+2)}=\frac{C_{0}}{2}-\frac{C_{1}}{3}+\frac{C_{2}}{4}-......+(-1)^{n}\frac{C_{n}}{n+2}$

So, Answer will be $D)$

Comments

@ankitgupta.1729

both side K addition will be better, because we r integrating both sides

right?

Then 3rd K will be merge of previous K's.

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yes mam, you can edit this thing if you want :)...you have written $K_1$ and $K_2$ and have not mentioned what is $k$ that's why I have written only $k$ to make it simple..

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ok :)