Question

$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}+\frac{C_3}{5}+..........$

• A. $\frac{1}{(n-1)(n+1)}$
• B. $\frac{1}{(n+1)(n+2)}$
• C. $\frac{1}{2(n-1)(n+1)}$
• D. None of these

Answer 1

The correct option is B $\frac{1}{(n+1)(n+2)}$

Integrating the expansion of $x{(1-x)}^n$ between the limits $0$ and $1$

${\int }_0^{1}x{(1-x)}^{n}dx$

$={\int }_0^{1}x(C_0-C_{1}x+C_2{x}^2-...+{(-1)}^n{C}_n{x}^n)dx$

$=C_0{\left[\frac{x^2}{2}\right]}_0^1-C_1{\left[\frac{x^3}{3}\right]}_0^1+C_2{\left[\frac{x^4}{4}\right]}_0^1-...$ .........$\left(1\right)$

The integral on LHS of eqn$\left(1\right)$

${\int }_1^0(1-t)t^n(-dt)$ by putting $1-x=t$

$={\int }_1^0(t^n-t^{n+1})dt$

$=\frac{1}{n+1}-\frac{1}{n+2}$

Whereas the integral on the of Eq.$\left(1\right)$

$\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-...to(n+1)$ terms

$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{n+2-n-1}{(n+1)(n+2)}$

$=\frac{1}{(n+1)(n+2)}$