## Discrete Mathematics with Applications 4th Edition

$\Sigma^{5}_{k=0}k(k-1)$ =$\Sigma^{6}_{i=1}(i-1)(i-2)$
When k = 0, then i = 1. When k = 5, then i = 6. Since i = k + 1, then k = i − 1. Thus, $k(k − 1) = (i − 1)((i − 1) − 1) = (i − 1)(i − 2),$ and so $\Sigma^{5}_{k=0}k(k-1)$ =$\Sigma^{6}_{i=1}(i-1)(i-2)$