Chapter 11.2, Problem 13CYU

To determine

Find the sum of ${\displaystyle \sum _{k=1}^{k=12}3k+9}$

Answer to Problem 13CYU

Option C is the correct choice.

Explanation of Solution

Given:

Find the sum of ${\displaystyle \sum _{k=1}^{k=12}3k+9}$

Concept Used:

K = 1 should be the first term and k = 12 should be the last term.

Calculation:

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{k=12}3k+9};k=1\\ 3k+9=3·1+9=12\\ a_1=12\end{array}$

  $\begin{array}{l}{\displaystyle \sum _{k=1}^{k=12}3k+9};k=12\\ 3k+9=3·12+9=45\\ a_{12}=45\end{array}$

  $\begin{array}{l}{a}_1=12;a_{12}=45,n=12\\ {\text{S}}_{12}=\frac{n}{2}(a_9+a_{21})\\ {\text{S}}_{12}=\frac{12}{2}(12+45)\\ {\text{S}}_{12}=6·57=342\end{array}$

Thus, Option C is the correct choice.