# How do you find the sum given Sigma k(k-2) from k=3 to 6?

Oct 31, 2016

${\sum}_{k = 3}^{k = 6} k \left(k - 2\right) = 50$
${\sum}_{k = 3}^{k = 6} k \left(k - 2\right) = 3 \left(3 - 2\right) + 4 \left(4 - 2\right) + 5 \left(5 - 2\right) + 6 \left(6 - 2\right)$
$\therefore {\sum}_{k = 3}^{k = 6} k \left(k - 2\right) = 3 \left(1\right) + 4 \left(2\right) + 5 \left(3\right) + 6 \left(4\right)$
$\therefore {\sum}_{k = 3}^{k = 6} k \left(k - 2\right) = 3 + 8 + 15 + 24$
$\therefore {\sum}_{k = 3}^{k = 6} k \left(k - 2\right) = 50$