Question

If $c_0,c_1,c_2,.......c_n$ denote the coefficients in the expansion of ${(1+x)}^n$, prove that
$\frac{c_1}{c_0}+\frac{2c_2}{c_1}+\frac{3c_3}{c_2}+...\frac{n{c}_n}{c_{n-1}}=\frac{n(n+1)}{2}$.

Answer 1

$S_n=\frac{c_1}{c_0}+\frac{2c_2}{c_1}+\frac{3c_3}{c_2}+\dots \dots \frac{n{c}_n}{c_{n-1}}$

$T_r=r\frac{c_r}{c_{r-1}}=r\cdot \frac{n!}{(n-r)!r!}\cdot \frac{(n-r+1)!(r-1)!}{n!}=n+1-r$

$R.H.S.=S_n=\sum _{r=1}^n{T}_n=\sum _{r=1}^n(n+1-r)=n^2+n-\frac{n(n+1)}{2}=\frac{n(n+1)}{2}=L.H.S.$