Question

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

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

Answer 1

The correct option is D none of these

${(1-x)}^n=C_0-C_{1}x+C_2{x}^2-C_3{x}^3+....$

$\therefore x{(1-x)}^n={xC}_0-C_1{x}^2+C_2{x}^3+....$

Integrate on both sides w.r.t. $x$ to get,

${\int }_0^{1}x{(1-x)}^{n}dx={\int }_0^1({xC}_0-C_1{x}^2+C_2{x}^3+....)dx=\frac{C_0}{2}-\frac{C_1}{3}+\frac{C_2}{4}...(n+1)terms$

For LHS put $1-x=t$, thus $dx=-dt$

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