# Expanding summation $\sum_{i=1}^{k+1}i(i!)$

Expand the summation: $\sum_{i=1}^{k+1}i(i!)$

My solution is:

$\sum_{i=1}^{k}i(i!)+k(k+1)$

Write $i(i!)$ as $(i+1-1)i!=(i+1)!-i!$ before summing.