Question

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

Answer 1

We know,

$(1+x{)}^n=c_0+c_{1}x+c_2{x}^2+...+c_n{x}^n$

Integrating both sides from $0$ to $1$

$\Rightarrow$ ${\int }_0^1(1+x{)}^n={\int }_0^1(c_0+c_{1}x+c_2{x}^2+...+c_n{x}^n)$

$\Rightarrow$ ${\left[\frac{(1+x{)}^{n+1}}{(n+1)}\right]}_0^1={[c_{0}x+c_1\frac{x^2}{2}+c_2\frac{x^3}{3}+...+c_n\frac{x^{n+1}}{(n+1)}]}_0^1$

$\Rightarrow$ $\frac{2^{n+1}}{n+1}-\frac{1}{(n+1)}=c_0+\frac{c_1}{2}+\frac{c_2}{3}+...+\frac{c_n}{n+1}$

$\Rightarrow$ $\frac{2^{n+1}-1}{n+1}=c_0+\frac{c_1}{2}+\frac{c_2}{3}+...+\frac{c_n}{n+1}$ --- Hence proved.