Question

The sum $\frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}-\frac{C_3}{4.5}+...$ to $(n+1)$ terms is

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

Answer 1

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

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

Integrating equation (1) w.r. to $x,$ between limits $0$ and $x,$we get

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

$\Rightarrow C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+...=\frac{{(1+x)}^{n+1}-1}{n+1}$ ...(2)

Integrating equation (2), taking limits from $-1$ to $0,$ we get

${\int }_{-1}^0[C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+...]dx={\int }_{-1}^0\frac{{(1+x)}^{n+1}-1}{n+1}dx.$ ...(3)

$\Rightarrow {[\frac{C_0{x}^2}{2}+\frac{C_1{x}^3}{2.3}+\frac{C_2{x}^4}{3.4}+...]}_{-1}^0={[\frac{{(1+x)}^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}]}_{-1}^0$

$\Rightarrow -[\frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}-...]$

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

$\therefore \frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}+...=\frac{1}{n+2}$