Sigma notation for an alternating sum

Since the sign is negative when $i$ is odd and positive when $i$ is even, the factor $(-1)^i$ produces exactly this pattern: $(-1)^1=-1$, $(-1)^2=1$, and so on. So each term can be written as $(-1)^i \cdot i$, giving $\sum_{i=1}^{9} (-1)^i \, i.$

To evaluate, group terms in pairs: $(-1+2)+(-3+4)+(-5+6)+(-7+8) = 1+1+1+1 = 4$, leaving the final unpaired term $-9$. Adding gives $4 - 9 = -5$, so the sum equals $-5$.

The leftover unpaired term was negative. After pairing the first eight terms into four groups of plus one each, the final term, negative nine, was left over by itself. This negative nine is what brought the running total of four down to negative five.