Chapter 12.1, Problem 78AYU

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

${\displaystyle \underset{k=8}{\overset{40}{\sum }}\left(-3k\right)}$

To determine

To find: The sum of given sequence: ${\displaystyle \sum }_{k=\text{ 8}}^{40}\left(-3k\right)$.

Answer to Problem 78AYU

${\displaystyle \sum }_{k=\text{ 8}}^{40}\left(-3k\right)=\text{ 2376}$

Explanation of Solution

Given:

${\displaystyle \sum }_{k=\text{ 8}}^{40}\left(-3k\right)$

Calculation:

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

${\displaystyle \sum }_{k=\text{ 8}}^{40}\left(-3k\right)=-\text{3}{\displaystyle \sum }_{k=\text{ 1}}^{40}\left(k\right)+\text{ 3}{\displaystyle \sum }_{k=\text{ 1}}^7\left(k\right)=\text{ 3}\left[\frac{40\left(40+1\right)}{2}\right]+\text{ 3}\left[\frac{7\left(\text{7 }+1\right)}{2}\right]$

${\displaystyle \sum }_{k=\text{ 8}}^{40}\left(-3k\right)=-2460+\text{ 84 }=\text{ 2376}$