## Calculus (3rd Edition)

$$\sum_{i=2}^{5} (i^2+i)$$
Given $$\left(2^{2}+2\right)+\left(3^{2}+3\right)+\left(4^{2}+4\right)+\left(5^{2}+5\right)$$ The first term is $\left(2^{2}+2\right)$, the last term is $\left(5^{2}+5\right)$, and we observe that the power is fixed at $2$; thus $$\left(2^{2}+2\right)+\left(3^{2}+3\right)+\left(4^{2}+4\right)+\left(5^{2}+5\right)=\sum_{i=2}^{5} (i^2+i)$$