Chapter 8.2, Problem 50E

To determine

To calculate: The sum ${\displaystyle \sum _{i=1}^{31}(-3-4i)}$ .

Answer to Problem 50E

The sum is $-2,077.$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^{31}(-3-4i)}$

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 ,

  ${\displaystyle \sum _{i=1}^{31}(-3-4i)}$

Step1. Calculating the first and last terms where number of terms $n=31.$

  $\begin{array}{l}{a}_i=-3-4i\\ a_1=-3-4(1)\\ =-3-4\\ =-7\\ a_{31}=-3-4(31)\\ =-3-124\\ =-127\end{array}$

Step2. Calculating the sum .

  $\begin{array}{l}{S}_{31}=31\left(\frac{(-7)+(-127)}{2}\right)\\ =31\left(\frac{-7-127}{2}\right)\\ =31\left(\frac{-134}{2}\right)\\ =31(-67)\\ =-2,077\end{array}$

Hence, the sum is $-2,077.$