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

$(1+x{)}^{n+1}=1+(n+1)x+\frac{(n+1)n}{1\cdot 2}{x}^2+\frac{(n+1)n(n-1)}{1\cdot 2\cdot 3}{x}^3+\dots \dots +(n+1)x^n+x^{n+1}$

$⟹\frac{(1+x{)}^{n+1}-1}{n+1}=x+\frac{n}{1\cdot 2}{x}^2+\frac{n(n-1)}{1\cdot 2\cdot 3}{x}^3+\dots \dots +\frac{x^{n+1}}{n+1}$

$⟹\frac{(1+x{)}^{n+1}-1}{n+1}=x+\frac{c_1}{2}{x}^2+\frac{c_2}{3}{x}^3+\dots \dots +\frac{c_n{x}^{n+1}}{n+1}$

Substituting $x=1$, we get

$\frac{2^{n+1}-1}{n+1}=1+\frac{c_1}{2}+\frac{c_2}{3}+\dots \dots +\frac{c_n}{n+1}$

Hence Proved