Chapter 8.2, Problem 47E

To determine

To calculate: The sum ${\displaystyle \sum _{i=1}^{20}(2i-3)}$ .

Answer to Problem 47E

The sum is $360.$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^{20}(2i-3)}$

Formula used:

Arithmetic difference − the difference of consecutive term is constant. The constant difference is called the common difference and is denoted by d .

Therefore,

  $d=a_{n+1}-a_n;n\ge 1$ .

The $nth$ tem of an arithmetic sequence with first term $a_1$ and common difference $d$ is given by : $a_n=a_1+(n-1)d$ .

The sum of the first n terms of arithmetic series is

  $S_n=n\left(\frac{a_1+a_n}{2}\right).$

Calculation:

Consider ,

  ${\displaystyle \sum _{i=1}^{20}(2i-3)}$

Step1. Calculating the first and last terms where number of terms $n=20.$

  $\begin{array}{l}{a}_i=2i-3\\ a_1=2(1)-3\\ =2-3\\ =-1\\ a_{20}=2(20)-3\\ =40-3\\ =37\end{array}$

Step2. Calculating the sum .

  $\begin{array}{l}{S}_{20}=20\left(\frac{(-1)+37}{2}\right)\\ =20\left(\frac{-1+37}{2}\right)\\ =20\left(\frac{36}{2}\right)\\ =20\left(18\right)\\ =360\end{array}$

Hence, the sum is $360.$