Question

The sum of the series,
$\frac{1}{2.3}\cdot 2+\frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+......$ to n terms is _____.

• A. $\frac{2^{n+1}}{n+2}+1$
• B. $\frac{2^{n+1}}{n+2}-1$
• C. $\frac{2^{n+1}}{n+2}+2$
• D. $\frac{2^{n+1}}{n+2}-2$

Answer 1

The correct option is B $\frac{2^{n+1}}{n+2}-1$

Let $S=\frac{1}{2.3}\cdot 2+\frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+......$

General term is given by, $T_r=\frac{r}{(r+1)(r+2)}\cdot 2^r=\frac{r\cdot 2^r}{(r+1)(r+2)}=\frac{2(r+1)-(r+2)}{(r+1)(r+2)}\cdot 2^r=\frac{(r+1)2^{r+1}-(r+2)2^r}{(r+1)(r+2)}$

$\Rightarrow T_r=\frac{2^{r+1}}{r+2}-\frac{2^r}{r+1}$, split the terms

Therefore the required sum $=\sum _{r=1}^n{T}_r=T_1+T_2+T_3+...........+T_n$

$=(\frac{2^2}{3}-\frac{2}{2})+(\frac{2^3}{4}-\frac{2^2}{3})+(\frac{2^4}{5}-\frac{2^3}{4})+..............+(\frac{2^{n+1}}{n+2}-\frac{2^n}{n+1})$

$=-\frac{2}{2}+\frac{2^{n+1}}{n+2}=\frac{2^{n+1}}{n+2}-1$, all the remaining terms will get cancelled