Let's find the sum of the series:
[tex]\sum ^{11}_{n\mathop=1}(3+2n)[/tex]
We'll use these formulas
[tex]\begin{gathered} \sum ^n_1K=K\cdot n \\ \sum ^{i=n}_{i=1}i=\frac{n(n+1)}{2} \end{gathered}[/tex]
We separate the series in two:
[tex]S_1=\sum ^{11}_{n\mathop=1}3=3\cdot11=33[/tex][tex]S_2=\sum ^{i=11}_{i\mathop=1}2\cdot i=2\sum ^{i=11}_{i\mathop{=}1}i=2\cdot\frac{11\cdot(11+1)}{2}=11\cdot12=132[/tex]
Thus, our sum is
S = 33 + 132 = 165
