Direct expansion of an alternating sigma sum

Substitute each value of $k$ from 0 to 5 into $(-1)^k(k+1)$: for $k=0$, term is $1$; for $k=1$, term is $-2$; for $k=2$, term is $3$; for $k=3$, term is $-4$; for $k=4$, term is $5$; for $k=5$, term is $-6$.

Now add the six terms in order: $1 - 2 + 3 - 4 + 5 - 6$. Grouping in pairs, $(1-2) + (3-4) + (5-6) = -1 -1 -1 = -3$.