Question

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

A
2n+1n+2+1
B
2n+1n+21
C
2n+1n+2+2
D
2n+1n+22

Solution

## The correct option is B $$\dfrac{2^{n+1}}{n+2}-1$$Let  $$S = \displaystyle \frac{1}{2.3}\cdot 2+ \frac{2}{3.4}\cdot 2^2+\frac{3}{4.5}\cdot 2^3+ ......$$General term is given by, $$T_r = \dfrac{r}{(r+1)(r+2)}\cdot 2^r=\dfrac{r\cdot 2^r}{(r+1)(r+2)}=\dfrac{2(r+1)-(r+2)}{(r+1)(r+2)}\cdot 2^r=\dfrac{(r+1)2^{r+1}-(r+2)2^r}{(r+1)(r+2)}$$$$\Rightarrow T_r = \dfrac{2^{r+1}}{r+2}-\dfrac{2^r}{r+1}$$, split the terms Therefore the required sum $$=\displaystyle \sum_{r=1}^nT_r=T_1+T_2+T_3+...........+T_n$$$$\quad =\left( \dfrac{2^2}{3}-\dfrac{2}{2}\right)+\left( \dfrac{2^3}{4}-\dfrac{2^2}{3}\right)+\left(\dfrac{2^4}{5}-\dfrac{2^3}{4} \right)+..............+\left(\dfrac{2^{n+1}}{n+2} -\dfrac{2^n}{n+1}\right)$$$$=-\dfrac{2}{2}+\dfrac{2^{n+1}}{n+2}=\dfrac{2^{n+1}}{n+2}-1$$, all the remaining terms will get cancelled Mathematics

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