Chapter 11.2, Problem 57PPS

To determine

Find the sum of ${\displaystyle \sum _{k=1}^{k=16}4k-2}$

Answer to Problem 57PPS

Sum of the series is 512.

Explanation of Solution

Given:

Find the sum of ${\displaystyle \sum _{k=1}^{k=16}4k-2}$

Concept Used:

K = 1 should be the first term and k = 16 should be the last term.

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{k=16}4k-2};k=1\\ 4k-2=4·1-2=2\\ a_1=2\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{k=16}4k-2};k=16\\ 4k-2=4·16-2=62\\ a_{16}=62\end{array}$

  $\begin{array}{l}{a}_1=2;a_{16}=62,n=16\\ {\text{S}}_{16}=\frac{n}{2}(a_9+a_{21})\\ {\text{S}}_{16}=\frac{16}{2}(2+62)\\ {\text{S}}_{16}=8·64=512\end{array}$

Thus, sum of the series is 512.