Question

In Problems 13–16, find each sum.

$\sum _{k=1}^{30}\left(k^2+2\right)$

Short answer

The sum is$\sum _{k=1}^{30}(k^2+2)=9515$

Step-by-step solution

Step 1. Given information.

The summation notation of the sequence is:

$\sum _{k=1}^{30}\left(k^2+2\right)$

Step 2. Expand the summation by substituting values of k.

$$\begin{gathered} \sum _{k=1}^{30}\left(k^2+2\right)=\left(1^2+2\right)+\left(2^2+2\right)+\left(3^2+2\right)+.............\left({30}^2+2\right) \\ =\left(1^2+2^2+3^2+....{30}^2\right)+(2+2+2..........30times) \\ =2\times 30+\left(1^2+2^2+3^2+....{30}^2\right) \\ =60+\frac{30(30+1)(2.30+1)}{6} \\ =60+\frac{30\times 31(60+1)}{6} \\ =60+5\left(61\left)\right(31)\right) \\ =60+9455 \\ =9515 \end{gathered}$$