Chapter 11.3, Problem 47PPS

To determine

Find the sum of the series: ${\displaystyle \sum _{k=1}^{7}4}·{(-3)}^{k-1}$

Answer to Problem 47PPS

Sum of the series is ${\text{S}}_{\text{n}}=2188$

Explanation of Solution

Given:

Find the sum of the series: ${\displaystyle \sum _{k=1}^{7}4}·{(-3)}^{k-1}$

Concept Used:

K=1 is the first term. Number of terms = 7 − 1 + 1 = 7

Calculation:

Number of terms: n = 7

First term:$k=1;{\displaystyle \sum _{k=1}^{7}4}·{(-3)}^{k-1}\Rightarrow 4·{(-3)}^{1-1}=4·1=4$

Find the common difference:

The base of the exponential function is the common ratio, so $r=-3$

  $\begin{array}{l}{a}_1=4;r=-3;n=7;\text{Find}{\text{S}}_{\text{n}}\\ {\text{S}}_{\text{n}}=\frac{a_1(1-r^n)}{1-r}\\ {\text{S}}_{\text{n}}=\frac{4(1-{(-3)}^7)}{1-(-3)}\\ {\text{S}}_{\text{n}}=\frac{4(2188)}{4}=2188\\ {\text{S}}_{\text{n}}=2188\end{array}$

Thus, sum of the series is ${\text{S}}_{\text{n}}=2188$