Chapter 11.2, Problem 58PPS

To determine

Find the sum of ${\displaystyle \sum _{k=4}^{k=13}4k+1}$

Answer to Problem 58PPS

Sum of the series is 350.

Explanation of Solution

Given:

Find the sum of ${\displaystyle \sum _{k=4}^{k=13}4k+1}$

Concept Used:

K = 4 should be the first term and k = 13 should be the last term.

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{k=4}^{k=13}4k+1};k=4\\ 4k+1=4·4+1=17\\ a_4=17\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{k=4}^{k=13}4k+1};k=13\\ 4k+1=4·13+1=53\\ a_{13}=53\end{array}$

  $\begin{array}{l}{a}_4=17;a_{13}=53,n=10\\ {\text{S}}_{10}=\frac{n}{2}(a_4+a_{13})\\ {\text{S}}_{10}=\frac{10}{2}(17+53)\\ {\text{S}}_{10}=5·70=350\end{array}$

Thus, sum of the series is 350.