Question

Evaluate the integral ${\int }_0^1{x}^{n-1}(1-x{)}^{n}dx$ and hence find the sum of the series $\frac{C_0}{n}-\frac{C_1}{n+1}+\frac{C_2}{n+2}-(-1{)}^n\frac{C_n}{2n}=\frac{(n!)(n-1)!}{\left(2n\right)!}$

Answer 1

$I_0={\int }_0^1{x}^{n-1}(1-x^3{)}^{0}dx$
$={\int }_0^1{x}^{n-1}[C_0-C_{1}x+C_2{x}^2-\cdots +(-1{)}^n{C}_n{x}^n]dx$
$={\int }_0^1[C_0{x}^{n-1}-C_1{x}^n+C_2{x}^{n+1}+...+(-1{)}^n{C}_n{x}^{2n-1}]dx$
$={[C_0\frac{x^n}{n}-C_1\frac{x^{n+1}}{n+1}+C_2\frac{x^{n+2}}{n+2}-\cdots +(-1{)}^n{C}_n\frac{x^{2n}}{2n}]}_0^1$
$=\frac{C_0}{n}-\frac{C_1}{n+1}+\frac{C_2}{n+2}-\cdots +(-1{)}^n\frac{C_n}{2n}$ = L.H.S.
Again putting x = $si{n}^2\theta$
$\therefore dx=2sin\theta cos\theta d\theta$ and adjust the limits as 0 to $\pi /2$.
$I={\int }_0^{\pi /2}2sin{}^{2n-1}\theta co{s}^{2n+1}\theta .2sin\theta cos\theta d\theta$
$={\int }_0^{\pi /2}2si{n}^{2n-}\theta co{s}^{2n+1}\theta d\theta$
$=2\frac{r\left(\frac{2n-1+1}{2}\right)r\left(\frac{2n+1+1}{2}\right)}{2r\left(\frac{4n+2}{2}\right)}=\frac{rnr(n+1)}{r(2n+1)}$
$=\frac{(n-1)!n!}{\left(2n\right)!}$ as r (n + 1) = n!