1 Answer

Joshua Wade · Answer 1 · 2023-01-20T22:48:33+0000

Let's use the properties of the sum:

$\begin{gathered} \sum ^(11)_(i\mathop=1)(3i^2+17)=\sum ^(11)_{i\mathop{=}1}3i^2+\sum ^(11)_{i\mathop{=}1}17 \\ =3\sum ^(11)_{i\mathop{=}1}i^2+17\cdot11 \end{gathered}$

Therefore the answer is the last option.

To simplify this we use the properties listed below:

$\sum ^n_{i\mathop{=}1}na_i=n\sum ^n_{i\mathop{=}1}a_i$
$\sum ^n_{i\mathop{=}1}(a_i+b_i)=\sum ^n_{i\mathop{=}1}a_i+\sum ^n_{i\mathop{=}1}b_i$
$\sum ^n_{i\mathop{=}1}c=cn$

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