The sum of the series 21⋅2+52⋅321+103⋅422+174⋅523+…… upto n terms is equal

Tr=r2+1r(r+1)⋅2r−1=1+1r−2r+12r−1Sn=∑r=1n Tr=∑r=1n 2r−1+∑r=1n 2r−1r−2rr+1=2n−1+1−2nn+1=nn+12n

The sum of the series 21⋅2+52⋅321+103⋅422+174⋅523+…… upto n terms is equal

Tr=r2+1r(r+1)⋅2r−1=1+1r−2r+12r−1Sn=∑r=1n Tr=∑r=1n 2r−1+∑r=1n 2r−1r−2rr+1=2n−1+1−2nn+1=nn+12n

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The sum of the series $\frac{2}{1 \cdot 2} + \frac{5}{2 \cdot 3} 2^{1} + \frac{10}{3 \cdot 4} 2^{2} + \frac{17}{4 \cdot 5} 2^{3} + \dots \dots$ upto n terms is equal

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$\frac{n 2^{n}}{n + 1}$

$(\frac{n}{n + 1}) 2^{n} + 1$

$\frac{n 2^{n}}{n + 1} - 1$

$\frac{(n - 1) 2^{n}}{n + 1}$

answer is A.

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Detailed Solution

$T_{r} = \frac{(r^{2} + 1)}{r (r + 1)} \cdot 2^{r - 1} = (1 + \frac{1}{r} - \frac{2}{r + 1}) 2^{r - 1}$
$S_{n} = \sum_{r = 1}^{n} T_{r} = \sum_{r = 1}^{n} 2^{r - 1} + \sum_{r = 1}^{n} (\frac{2^{r - 1}}{r} - \frac{2^{r}}{r + 1}) = (2^{n - 1}) + [1 - \frac{2^{n}}{n + 1}] = (\frac{n}{n + 1}) 2^{n}$