Chapter 12.1, Problem 76AYU

In Problems 71-82, find the sum of each sequence.

${\displaystyle \underset{k=1}{\overset{26}{\sum }}\left(3k-7\right)}$

To determine

To find: The sum of given sequence: ${\displaystyle \sum }_{k=1}^{26}\left(3k-\text{ 7}\right)$.

Answer to Problem 76AYU

${\displaystyle \sum }_{k=1}^{26}\left(3k-\text{ 7}\right)=\text{ 871}$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=1}^{26}\left(3k-\text{ 7}\right)$

Calculation:

Using ${\displaystyle \sum }_{\text{k }=1}^{\text{n}}k=1+2+3+\dots +\text{ n }=\frac{n\left(n+1\right)}{2}$ and ${\displaystyle \sum }_{\text{k }=1}^{\text{n}}\text{c }=\text{ cn}$ we have

${\displaystyle \sum }_{k=1}^{26}\left(3k-\text{ 7}\right)=\text{ 3}{\displaystyle \sum }_{k=1}^{26}k-{\displaystyle \sum }_{k=1}^{26}\text{7 }=\text{ 3}\left[\frac{26\left(26+1\right)}{2}\right]-26*\text{ 7 }=1053–182=871$