Berkeley Math Circle: Monthly Contest 7

Solution:
Observe that $3^{2}-2^{2}=(3-2)(3+2)=3+2$, $5^{2}-4^{2}=(5-4)(5+4)=5+4$, and so on. Thus this sum, call it $S$, is actually equal to
$$S=1+(2+3)+(4+5)+\cdots+(97+98)+99 .$$
We can also write it in reverse as
$$S=99+98+97+96+95+\cdots+3+2+1 .$$
Adding these two, we get
$$2 S=\underbrace{100+100+\cdots+100}_{99 \text{ times}}$$
Thus, $S=\frac{1}{2} \cdot 100 \cdot 99=4950$.