Question

$4C_0+\frac{4^2}{2}.C_1+\frac{4^3}{3}{C}_2+............+\frac{4^{n+1}}{n+1}{C}_n=\frac{5^{n+1}-1}{n+1}$

Answer 1

We have :

${(1+x)}^n=n_{C_0}+n_{C_1}x+n_{C_2}{x}^2+n_{C_3}{x}^3+......+n_{C_n}{x}^n$

Integrating both sides under the limits $[4,0]$ we get

${\int }_0^4{(1+x)}^{n}dx=n_{C_0}{\int }_0^{4}dx+n_{C_1}{\int }_0^{4}xdx+n_{C_2}{\int }_0^4{x}^{2}dx+......+n_{C_n}{\int }_0^4{x}^{n}dx$

$\frac{1+x^{n+1}}{n+1}{\int }_0^4=n_{C_0}\left(4\right)+n_{C_1}\frac{{\left(4\right)}^2}{2}+n_{C_2}\frac{{\left(4\right)}^3}{3}+......+n_{C_n}\frac{4^{n+1}}{n+1}$

$\frac{5^{n+1}-1}{n+1}=n_{C_0}\left(4\right)+n_{C_1}\frac{\left(4\right)}{2}+n_{C_2}\frac{{\left(4\right)}^3}{3}+......+n_{C_n}\frac{4^{n-x}}{n+1}$

Hence proved.