2. By how much is the sum of the squares of the first hundred even numbers greater than the sum of the squares of the first hundred odd numbers
Answer: 20100.

Solution: Group the terms as $\left(2^{2}-1^{2}\right)+\left(4^{2}-3^{2}\right)+\cdots+\left(200^{2}-199^{2}\right)=$ $(2-1) \cdot(2+1)+(4-3) \cdot(4+3)+\ldots+(200-199) \cdot(200+199)=1+2+\cdots+$ $199+200$. Divide the terms into pairs that give the same sum: $1+200=2+199=\ldots=100+101=201$. There will be exactly 100 such pairs, so the sum is 20100.