Question

If ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+....+C_n{x}^n$, then find the sum of the series
$\frac{C_0}{2}-\frac{C_1}{6}+\frac{C_2}{10}-\frac{C_3}{14}+........+\frac{{(-1)}^n{C}_n}{4n+2}$

Answer 1

Given, $(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+......+C_n{x}^n$

Consider $\int (1-x^2{)}^{n}dx=\int (C_0+C_1(-x^2)+C_2(-x^2{)}^2+.....+(-1{)}^n{C}_n{x}^{2n})$

$\Rightarrow \int (1-x^2{)}^{n}dx=(C_0-\frac{C_1{x}^3}{3}+.......+\frac{(-1{)}^n{C}_n{x}^{2n+1}}{2n+1})$

Substitute $x=1$ in above equation:

Required sum of series is $\frac{\int (1-x^2{)}^{n}dx}{2}$ at $x=1$.