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Catskul · Answer 1 · 2023-01-31T07:02:14+0000

Let's find the sum of the series:

$\sum ^(11)_(n\mathop=1)(3+2n)$

We'll use these formulas

$\begin{gathered} \sum ^n_1K=K\cdot n \\ \sum ^(i=n)_(i=1)i=(n(n+1))/(2) \end{gathered}$

We separate the series in two:

$S_1=\sum ^(11)_(n\mathop=1)3=3\cdot11=33$
$S_2=\sum ^(i=11)_(i\mathop=1)2\cdot i=2\sum ^(i=11)_{i\mathop{=}1}i=2\cdot(11\cdot(11+1))/(2)=11\cdot12=132$

Thus, our sum is

S = 33 + 132 = 165

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