HMMT November

Note that $(n+3)^{2}-(n+2)^{2}-(n+1)^{2}+n^{2}=4$ for every $n$. Therefore, adding $0^{2}$ to the end of the given sum and applying this identity for every four consecutive terms after $100^{2}$, we see that the given sum is equivalent to $100^{2}+25 \cdot 4=10100$. Alternatively, we can apply the difference-of-squares factorization to rewrite $100^{2}-98^{2}=(100-98)(100+98)=2(100+98), 99^{2}-97^{2}=(99-97)(99+97)=2(99+97)$, etc. Thus, the given sum is equivalent to $2(100+99+\cdots+2+1)=2 \cdot \frac{100 \cdot 101}{2}=10100$