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