The sum of n terms of the series 1+21+1n+31+1n2+.... is

a=1 d=2-1=1 r=1+1n S∞=a1-r+dr(1-r)2=11-1-1n+(1)1+1n1-1-1n2 =-n+1+1n1n2=-n+(n+1) n =-n+n2+n=n2

The sum of n terms of the series 1+21+1n+31+1n2+.... is

a=1 d=2-1=1 r=1+1n S∞=a1-r+dr(1-r)2=11-1-1n+(1)1+1n1-1-1n2 =-n+1+1n1n2=-n+(n+1) n =-n+n2+n=n2

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The sum of n terms of the series $1 + 2 (1 + \frac{1}{n}) + 3 {(1 + \frac{1}{n})}^{2} + . . . .$ is

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

$n^{2}$

$n (n + 1)$

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

answer is A.

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Detailed Solution

$a = 1 d = 2 - 1 = 1 r = 1 + \frac{1}{n} S \infty = \frac{a}{1 - r} + \frac{dr}{{(1 - r)}^{2}} = \frac{1}{1 - 1 - \frac{1}{n}} + \frac{(1) (1 + \frac{1}{n})}{{(1 - 1 - \frac{1}{n})}^{2}} = - n + \frac{1 + \frac{1}{n}}{\frac{1}{n^{2}}} = - n + (n + 1) n = - n + n^{2} + n = n^{2}$