Given:
The sum is,
[tex]\sum ^4_{i=1}\frac{5(4i-1)^2}{4}[/tex]
Expand the sum,
[tex]\begin{gathered} \sum ^4_{i=1}\frac{5(4i-1)^2}{4}=\frac{5(4(1)-1)^2}{4}+\frac{5(4(2)-1)^2}{4}+\frac{5(4(3)-1)^2}{4}+\frac{5(4(4)-1)^2}{4} \\ =\frac{5(3)^2}{4}+\frac{5(7)^2}{4}+\frac{5(11)^2}{4}+\frac{5(15)^2}{4} \\ =\frac{45}{4}+\frac{245}{4}+\frac{605}{4}+\frac{1125}{4} \\ =\frac{2020}{4} \end{gathered}[/tex]
Answer: option b)
[tex]\frac{2020}{4}[/tex]
