Chapter 12.2, Problem 49AYU

In Problems 39-56, find each sum.

${\displaystyle \underset{n=1}{\overset{80}{\sum }}\left(2n-5\right)}$

To determine

To find: The sum of the given sequence.

Answer to Problem 49AYU

$6080$

Explanation of Solution

Given:

${\displaystyle \sum }_{n=1}^{80}\left(2n-5\right)$

Formula Used:

Sum of the first $n$ terms of an arithmetic sequence.

$S_n=\frac{n}{2}\left[a_1+a_n\right]$

Calculation:

From the given sequence, $a_1=-3,n=80\text{ and }{a}_{80}=155$.

To find the sum ${\displaystyle \sum }_{n=1}^{80}\left(2n-5\right)$:

${\displaystyle \sum }_{n=1}^{80}\left(2n-5\right)=\frac{80}{2}\left[-3+155\right]=40\times 152=6080$   (By Formula)