Summing up squares

Solution 1:

$\begin{aligned}\sum\limits_{n=0}^{99} (n+1)^2 &= \sum\limits_{n=1}^{100} (n)^2 \\ \\ &= \frac{(n)(n+1)(2n+1)}{6} \\ \\ &= \frac{(100)(101)(201)}{6} \\ \\ &= \boxed{338350} \end{aligned}$

Solution 2:

$\begin{aligned}\sum\limits_{n=0}^{99} (n+1)^2 &= \sum\limits_{n=0}^{99} n^2 + 2n + 1 \\ \\ &= \frac{(n)(n+1)(2n+1)}{6} + (2) \dfrac{(n)(n+1)}{2} + (n+1) \\ \\ &= \frac{(99)(100)(199)}{6} + (99)(100) + (100) \\ \\ &= \boxed{338350} \end{aligned}$