Question

The sum to (n+1) terms of the following series $\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\frac{C_3}{5}+\cdots$ is

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

Answer 1

The correct option is D None of these

$\Rightarrow x(1-x{)}^n=C_{0}x-C_1{x}^2+C_2{x}^3-C_3{x}^4+\cdots$
$\Rightarrow {\int }_0^{1}x(1-x{)}^{n}dx={\int }_0^1(C_{0}x-C_1{x}^2+C_2{x}^3\cdots )dx\cdots (i)$
The integral on the LHS
$={\int }_1^0(1-t)t^n(-dt),byputting1-x=t$
$={\int }_0^1(t^n-t^{n+1})dt=\frac{1}{n+1}-\frac{1}{n+2}$
whereas the integral on the RHS of (i)
$=[\frac{C_0{x}^2}{2}-\frac{C_1{x}^3}{3}+\frac{C_2{x}^4}{4}-\cdots ]=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\cdots$
$\therefore \frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}-\cdots$ to (n+1) terms
$=\frac{1}{n+1}-\frac{1}{n+2}=\frac{1}{(n+1)(n+2)}$