Question

Sol. [Key Concept: The $n$-th term is $T_{n} = \frac{1}{n}{(1}^{2} + dots + n^{2}) = \frac{(n + 1)(2n + 1)}{6}$; sum $T_{n}$ for $n = 1$ to $10$.]

$${T_{n} = \frac{1}{n}\left( 1^{2} + 2^{2} + \ldots + n^{2} \right) }{T_{n} = \frac{n(n + 1)(2n + 1)}{6n} }{T_{n} = \frac{1}{6}\left\lbrack 2n^{2} + 3n + 1 \right\rbrack }{S_{n} = \Sigma T_{n} = \frac{1}{6}\left\lbrack 2\Sigma n^{2} + 3\Sigma n + \Sigma 1 \right\rbrack }{\Sigma T_{n} = \frac{1}{6}\left\lbrack 2 \cdot \frac{n(n + 1)(2n + 1)}{6} + \frac{3n(n + 1)}{2} + n \right\rbrack }{S_{10} = \frac{1}{6}\left\lbrack \frac{2 \times 10 \times 11 \times 21}{6} + \frac{3}{2} \times 10 \times 11 + 10 \right\rbrack }{= \frac{10}{6}\left\lbrack 77 + \frac{33}{2} + 1 \right\rbrack = \frac{315}{2} }$$