Let's use the properties of the sum:
[tex]\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}[/tex]
Therefore the answer is the last option.
To simplify this we use the properties listed below:
[tex]\sum ^n_{i\mathop{=}1}na_i=n\sum ^n_{i\mathop{=}1}a_i[/tex][tex]\sum ^n_{i\mathop{=}1}(a_i+b_i)=\sum ^n_{i\mathop{=}1}a_i+\sum ^n_{i\mathop{=}1}b_i[/tex][tex]\sum ^n_{i\mathop{=}1}c=cn[/tex]
