Chapter 8.2, Problem 49E

To determine

To calculate: The sum ${\displaystyle \sum _{i=1}^{33}(6-2i)}$ .

Answer to Problem 49E

The sum is $-924.$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^{33}(6-2i)}$

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 ,

  ${\displaystyle \sum _{i=1}^{33}(6-2i)}$

Step1. Calculating the first and last terms where number of terms $n=33.$

  $\begin{array}{l}{a}_i=6-2i\\ a_1=6-2(1)\\ =6-2\\ =4\\ a_{33}=6-2(33)\\ =6-66\\ =-60\end{array}$

Step2. Calculating the sum .

  $\begin{array}{l}{S}_{33}=33\left(\frac{4+(-60)}{2}\right)\\ =33\left(\frac{4-60}{2}\right)\\ =33\left(\frac{-56}{2}\right)\\ =33(-28)\\ =-924\end{array}$

Hence, the sum is $-924.$