Question

Prove that
$\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+....=\frac{2n}{n+1}$

Answer 1

Consider the given function:

$\frac{c_0}{1}+\frac{c_2}{3}+\frac{c_4}{5}+......+=\frac{2^n}{n+1}$

$L.H.S$

$=\frac{1}{1}+\frac{n(n-1)}{2!3}+\frac{n(n-1)(n-2)(n-3)}{4!5}+......$

$=\frac{1}{n+1}\left[\right(n+1)+\frac{n(n-1)(n+1)}{3!}+\frac{n(n-1)(n-2)(n-3)(n+1)}{5!}......$

$putthan(n+1)=m$

$n=(m-1)$

$=\frac{1}{m}[m+\frac{m(m-1)(m-2)}{3!}+\frac{m(m-1)(m-2)(m-3)}{5!}......]$

$=\frac{1}{m}[m+m{c}_3+m{c}_5+......]$

$=\frac{1}{m}[m{c}_1+m{c}_3+m{c}_5+.......]$

$=\frac{1}{m}\left[2^{(m-1)}\right]$

$=\frac{2n}{n+1}R.H.S$

Hence this is the answer.

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