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Question: The sum to n terms of the series \(\frac{3}{1^{2}}\) + \(\frac{5}{1^{2} + 2^{2}}\) + \(\frac{7}{1^{2...

The sum to n terms of the series 312\frac{3}{1^{2}} + 512+22\frac{5}{1^{2} + 2^{2}} + 712+22+32\frac{7}{1^{2} + 2^{2} + 3^{2}} + …… is

A

3nn+1\frac{3n}{n + 1}

B

6nn+1\frac{6n}{n + 1}

C

9nn+1\frac{9n}{n + 1}

D

12nn+1\frac{12n}{n + 1}

Answer

6nn+1\frac{6n}{n + 1}

Explanation

Solution

Tn = (2n+1)12+22+...+n2\frac{(2n + 1)}{1^{2} + 2^{2} + ... + n^{2}} = 6(2n+1)n(n+1)(2n+1)\frac{6(2n + 1)}{n(n + 1)(2n + 1)}

= 6n(n+1)\frac{6}{n(n + 1)} = 6 [1n–1n+1]\left\lbrack \frac{1}{n}–\frac{1}{n + 1} \right\rbrack

\ Sn = T1 + T2 + ….. + Tn

= 6 [1–12+12–13+13–14+.....+1n–1n+1]\left\lbrack 1–\frac{1}{2} + \frac{1}{2}–\frac{1}{3} + \frac{1}{3}–\frac{1}{4} + ..... + \frac{1}{n}–\frac{1}{n + 1} \right\rbrack= 6nn+1\frac{6n}{n + 1}