Chapter F, Problem 21E

To determine

To find: The value of the sum ${\displaystyle \sum _{i=4}^8\left(3i-2\right)}$.

Answer to Problem 21E

The value of the sum ${\displaystyle \sum _{i=4}^8\left(3i-2\right)}$ is $80$.

Explanation of Solution

Definition used:

If $a_m,a_{m+1},\mathrm{...},a_n$ are real numbers and m and n are integers such that $m\le n$, then ${\displaystyle \sum _{i=m}^n{a}_i=a_m+}{a}_{m+1}+a_{m+2}+\cdots +a_{n-1}+a_n$.

Calculation:

By the above definition, the sum ${\displaystyle \sum _{i=4}^8\left(3i-2\right)}$ expressed as follows.

$\begin{array}{c}{\displaystyle \sum _{i=4}^8\left(3i-2\right)}=\left(3\left(4\right)-2\right)+\left(3\left(5\right)-2\right)+\left(3\left(6\right)-2\right)+\left(3\left(7\right)-2\right)+\left(3\left(8\right)-2\right)\\ =\left(12-2\right)+\left(15-2\right)+\left(18-2\right)+\left(21-2\right)+\left(24-2\right)\\ =10+13+16+19+22\\ =80\end{array}$

Thus value of the sum ${\displaystyle \sum _{i=4}^8\left(3i-2\right)}$ is $80$.