How do you find the sum of the series $k^2$ from k=1 to 35?

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Answer 1 · 2016-11-10T13:56:51

$sum_(k=1)^35 k^2 = 14910$

We need the standard formula $sum_(r=1)^n r^2=1/6n(n+1)(2n+1)$

$:. sum_(k=1)^35 k^2 = 1/6(35)(35+1)(70+1)$
$:. sum_(k=1)^35 k^2 = 1/6(35)(36)(71)$
$:. sum_(k=1)^35 k^2 = 14910$

How do you find the sum of the series k^2k^2 from k=1 to 35?