Alternating sum of squares from 100 down

Starting at $100^2$, alternately subtract and add the squares of each smaller positive integer, ending by subtracting $1^2$. In other words, evaluate $\sum_{k=1}^{50}\left[(2k)^2-(2k-1)^2\right].$

5050. Pair consecutive terms: $a^2-(a-1)^2=a+(a-1)$. The sum therefore equals $100+99+98+\cdots+1=\tfrac{100\cdot101}{2}=5050$.