Question

The sum $\frac{C_{0}}{1.2}-\frac{C_{1}}{2.3}+\frac{C_{2}}{3.4}-\frac{C_{3}}{4.5}+\ldots$ 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 $(1+x) n=C_{0}+C_{1} x+C_{2} x^{2}+C_{3} x^{3}+\ldots \ldots$....(1)
Integrating equation (1) w.r. to $x$, between limits 0 and $x$, we get
$\int\limits_{0}^{x}\left(C_{0}+C_{1} x+C_{2} x^{2}+\ldots \ldots\right) d x=\int\limits_{0}^{x}(1+x)^{n} d x$
$\Rightarrow C_{0} x+C_{1} \frac{x^{2}}{2}+C_{2} \frac{x^{3}}{3}+\ldots . .=\frac{(1+x)^{n+1}-1}{n+1} ....$(2)
Integrating equation (2), taking limits from $-1$ to $0$ , we get
$\int\limits_{-1}^{0}\left[C_{0} x+C_{1} \frac{x^{2}}{2}+C_{2} \frac{x^{3}}{3}+\ldots\right] d x=\int\limits_{-1}^{0} \frac{(1+x)^{n+1}-1}{n+1} d x ....$(3)
$\Rightarrow \left[\frac{C_{0} x^{2}}{2}+\frac{C_{1} x^{3}}{2.3}+\frac{C_{2} x^{4}}{3.4}+\ldots .\right]_{-1}^{0}=\left[\frac{(1+x)^{n+2}}{(n+1)(n+2)}-\frac{x}{n+1}\right]_{-1}^{0}$
$\Rightarrow -\left[\frac{C_{0}}{1.2}-\frac{C_{1}}{2.3}+\frac{C_{2}}{3.4}-\ldots . .\right.$
$=\frac{1}{(n+1)(n+2)}-\frac{1}{n+1}=-\frac{1}{n+2}$
$\therefore \frac{C_{0}}{1.2}-\frac{C_{1}}{2.3}+\frac{C_{2}}{3.4}+\ldots .=\frac{1}{n+2}$