Answer
$\frac{1}{3}n(4n^2+24n+47)$
Work Step by Step
Expanding the expression under the sum and then summing term by term, with formulas from theorem 3, we get, step by step:
$$A=\sum_{i=1}^n(3+2i)^2=\sum_{i=1}^n(9+12i+4i^2)=\\ =\sum_{i=1}^n9+12\sum_{i=1}^ni+4\sum_{i=1}^ni^2=\\=
9n+12\frac{n(n+1)}{2}+4\frac{n(n+1)(2n+1)}{6}=\\=
9n+6n(n+1)+\frac{2}{3}n(n+1)(2n+1)=\\=
n\left(9+6(n+1)+\frac{2}{3}(n+1)(2n+1)\right)=\\=
n\left(9+(n+1)\left(6+\frac{2}{3}(2n+1)\right)\right)=\\=
n\left(9+(n+1)\left(\frac{4n+20}{3}\right)\right)=\\=
n\left(9+\frac{4(n+1)(n+5)}{3}\right)=\\=\frac{1}{3}n(4n^2+24n+47).$$