How do you evaluate the sum represented by $10 \sum n = 1 n^{2}$ $sum_(n=1)^(10)n^2$ ?

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Answer 1 · 2014-09-21T16:12:53

Since

$n \sum k = 1 k^{2} = \frac{n (n + 1) (2 n + 1)}{6}$ $sum_{k=1}^n k^2={n(n+1)(2n+1)}/{6}$ ,

we have

$10 \sum n = 1 n^{2} = \frac{(10) (11) (21)}{6} = 385$ $sum_{n=1}^{10}n^2={(10)(11)(21)}/6=385$

How do you evaluate the sum represented by sum_(n=1)^(10)n^210∑n=1n2sum_(n=1)^(10)n^2 ?