Question

The sum of the series,
$\frac{1}{2.3}.2+\frac{2}{3.4}.2^2+\frac{3}{4.5}.2^3+$ ............ 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

Answer: (B)

Given series is
$\frac{1}{2\cdot 3}\cdot 2^1+\frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+\dots$upto $n$ terms
The $n$ th term of the series is
$T_n=\frac{n}{(n+1)(n+2)}\cdot 2^n=\left\{\frac{2\cdot 2^n}{(n+2)}-\frac{2^n}{(n+1)}\right\}$
On putting $n=1,2,3\dots$, we get
$T_1=2\cdot \frac{2^1}{3}-\frac{2^1}{2},T_2=\frac{2\cdot 2^2}{4}-\frac{2^2}{3},T_3=\frac{2\cdot 2^3}{5}-\frac{2^3}{4}$
.................
.................
On adding, we get
$S_n=T_1+T_2+T_3+\dots +T_n$
$\Rightarrow S_n=\frac{2\cdot 2^n}{(n+2)}-\frac{2}{2}=\frac{2^{n+1}}{(n+2)}-1$