## Calculus 8th Edition

The solution is $$\sum_{i=1}^n(i^2+3i+4)=\frac{1}{3}n\left(n^2+6n+17\right).$$
First, we will use the fact that the sum passes through terms individually, as well as the fact that the constant can be put in front of the sum: $$A=\sum_{i=1}^n(i^2+3i+4)=\sum_{i=1}^ni^2+3\sum_{i=1}^ni+\sum_{i=1}^n4.$$ Now, using equalities from Theorem 3 we get $$A=\frac{n(n+1)(2n+ 1)}{6}+3\frac{n(n+1)}{2}+4n=\\ n\left(\frac{(n+1)(2n+1)}{6}+\frac{3(n+1)}{2}+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\left(\frac{2n+1}{6}+\frac{3}{2}\right)+4\right)=n\left((n+1)\frac{2n+1+9}{6}+4\right)=n\left(\frac{(n+1)(n+5)}{3}+4\right)=\frac{1}{3}n\left(n^2+6n+17\right).$$