Question

Find the sum of n terms of the series : n+1/n, n+3/n , n+5/n, .....

Answer 1

Given,
$a = n + \frac{1}{n} \\ \\ a = \frac{ {n}^{2} + 1}{n} \\ \\ a_{2} = n + \frac{3}{n} \\ a_{2} = \frac{ {n}^{2} + 3}{n} \\ \\ d = a_{2} - a_{1} \\ d = \frac{ {n}^{2} + 3}{n} - \frac{ {n}^{2} + 1}{n} \\ d = \frac{ {n}^{2} + 3 - {n}^{2} - 1}{n} \\ d = \frac{2}{n}$

Now,

$\bf \: S_{n} \: = \frac{n}{2} (2a + (n - 1)d) \\ \\ S_n = \frac{n}{2} (2( \frac{ {n}^{2} + 1 }{n} ) + (n - 1)( \frac{2}{n} ) \\ \\ S_n = \frac{n}{2} ( \frac{2 {n}^{2} + 2 }{n} ) + ( \frac{2n}{n} - \frac{2}{n} ) \\ \\ S_n = \frac{n}{2} ( \frac{2 {n}^{2} }{n} + \frac{2}{n} + \frac{2n}{n} - \frac{2}{n} ) \\ \\ S_n = \frac{n}{2} ( \frac{2 {n}^{2} }{n} + \frac{2n}{n} ) \\ \\ S_n = \frac{2n}{2n} ( {n}^{2} + n) \\ \\ S_n = {n}^{2} + n$

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