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If $(1+x)^n=C_0+C_1 x+C_2 x^2+\ldots .+C_n x^n$ , then $C_0 C_2+C_1 C_3+C_2 C_4+C_{n-2} C_n$ equals
Option: 1
$\frac{(2 n) !}{(n+1) !(n+2) !}$

Option: 2
$\frac{(2 n) !}{(n-2) !(n+2) !}$

Option: 3
$\frac{(2 n) !}{(n) !(n+2) !}$

Option: 4
$\frac{(2 n) !}{(n-1) !(n+2) !}$

Answers (1)

best_answer

We have, $(1+x)^n=C_0+C_1 x+C_2 x^2 \ldots+C_n x^n$

$\left(1+\frac{1}{x}\right)^n=C_0+C_1 \cdot \frac{1}{x}+C_2 \cdot \frac{1}{x^2}+\ldots+C_n\left(\frac{1}{x^n}\right)$

on multiplying both expansions, we get

$\begin{aligned} \frac{(1+x)^{2 n}}{x^n}=\sum C_0^2+x \sum C_0 C_1+x^2 \sum C_0 C_2+\ldots . \ +x^r \sum C_0 C_r+\ldots . \end{aligned}$

The various sigma are the coefficient of $x^0, x, x^2, \ldots \ldots, x^r$ in L.H.S. $\frac{(1+x)^{2 n}}{x^n}$ or coefficient of $x^n, x^{n+1}, x^{n+2}, \ldots \ldots, x^{n+r}$

in the expansion of $(1+x)^{2 n}$ which occur in $T_{n+1}, T_{n+2}, \ldots$ and are

${ }^{2 n} C_n,{ }^{2 n} C_{n+1},{ }^{2 n} C_{n+2} \ldots{ }^{2 n} C_{n+r} \text { etc. }$

${ }^{2 n} C_{n+2}=\frac{(2 n) !}{(n-2) !(n+2) !}$

Posted by

Suraj Bhandari

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