Chapter 14.7, Problem 7MRE

To determine

To write: The series using sigma notation and then find the sum.

Answer to Problem 7MRE

The given series in sigma notation is ${\displaystyle \sum _{n=0}^{19}\left(4n-5\right)}$ and its sum is $660$ .

Explanation of Solution

Given: $-5-1+3+\mathrm{...}+71$ .

Consider the given series.

  $-5-1+3+\mathrm{...}+71$

The given series is arithmetic with common difference $4$ .

The $n^{\text{th}}$ term of the AP is given as,

  $a_n=a_1+\left(n-1\right)d$

Substitute the value in the above expression.

  $\begin{array}{c}71=-5+\left(n-1\right)4\\ 71=-5+4n-4\\ n=20\end{array}$

The series in sigma notation is written as,

  ${\displaystyle \sum _{n=0}^{19}\left(4n-5\right)}$

The sum of the series is calculated as,

  $\begin{array}{c}{S}_n=\frac{n\left(a_1+a_n\right)}{2}\\ S_{20}=\frac{20\left(-5+71\right)}{2}\\ =660\end{array}$

Therefore, the given series in sigma notation is ${\displaystyle \sum _{n=0}^{19}\left(4n-5\right)}$ and its sum is $660$ .