Question

The sum $\frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}-\frac{C_3}{4.5}+\dots$ 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

Answer: (A)

Consider $(1+x)n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+\dots \dots$....(1)
Integrating equation (1) w.r. to $x$, between limits 0 and $x$, we get
$\underset{0}{\overset{x}{\int }}\left(C_0+C_{1}x+C_2{x}^2+\dots \dots \right)dx=\underset{0}{\overset{x}{\int }}(1+x{)}^{n}dx$
$\Rightarrow C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+\dots ..=\frac{(1+x{)}^{n+1}-1}{n+1}....$(2)
Integrating equation (2), taking limits from $-1$ to $0$ , we get
$\underset{-1}{\overset{0}{\int }}\left[C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+\dots \right]dx=\underset{-1}{\overset{0}{\int }}\frac{(1+x{)}^{n+1}-1}{n+1}dx....$(3)
$\Rightarrow {\left[\frac{C_0{x}^2}{2}+\frac{C_1{x}^3}{2.3}+\frac{C_2{x}^4}{3.4}+\dots .\right]}_{-1}^0={\left[\frac{(1+x{)}^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}\right]}_{-1}^0$
$\Rightarrow -[\frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}-\dots ..$
$=\frac{1}{(n+1)(n+2)}-\frac{1}{n+1}=-\frac{1}{n+2}$
$\therefore \frac{C_0}{1.2}-\frac{C_1}{2.3}+\frac{C_2}{3.4}+\dots .=\frac{1}{n+2}$