The sum up to10 terms is equal to: [2026]

Question

The sum $1 + \frac{1}{2} (1^{2} + 2^{2}) + \frac{1}{3} (1^{2} + 2^{2} + 3^{3}) + . . .$ up to10 terms is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

Given, $1 + \frac{1}{2} (1^{2} + 2^{2}) + \frac{1}{3} (1^{2} + 2^{2} + 3^{3}) . . .$ up to 10 terms

$T_{n} = \frac{1}{n} \sum_{k = 1}^{n} k^{2}$

$T_{n} = \frac{1}{n} [\frac{n (n + 1) (2 n + 1)}{6}]$ $[∵ \sum_{k = 1}^{n} k^{2} = \frac{n (n + 1) (2 n + 1)}{6}]$

$= \frac{(n + 1) (2 n + 1)}{6} = \frac{2 n^{2} + 3 n + 1}{6}$

$S_{10} = \sum_{n = 1}^{10} T_{n}$

$\Rightarrow S_{10} = \sum_{n = 1}^{10} \frac{2 n^{2} + 3 n + 1}{6}$

$S_{10} = \frac{1}{6} [\sum_{n = 1}^{10} n^{2} + 3 \sum_{n = 1}^{10} n + \sum_{n = 1}^{10} 1]$

$= \frac{1}{6} [2 \times \frac{(10 \times 11 \times 21)}{6} + 3 \times \frac{10 \times 11}{2} + 10]$

$= \frac{1}{6} [770 + 165 + 10] = \frac{315}{2}$

Want Us to Email you About Special Offers & Updates?