Sigma Rules

Writing out our sum we get:
$f(100)=1^2-2^2+3^2-4^2+...+99^2-100^2$
We can write this sum as follows where $1\leq i\leq 100$ :
$f(100)=\displaystyle \sum_{\text{i even}}\left( (i-1)^2-i^2\right)=\sum_{\text{i even}}(1-2i)$
Now, let $i=2k$ to obtain:
$f(100)=\displaystyle \sum_{k=1}^{50}(1-4k)=50-4\sum_{k=1}^{50}k=50-4\frac{50\cdot 51}{2}=50-50\cdot 102=50(1-102)=-50\cdot 101=\boxed{-5050}$
Note: Remember that: $\displaystyle \sum_{k=1}^{l}k=\frac{l(l+1)}{2}$