Chapter 8.3, Problem 47E

To determine

The sum of geometric progression.

Answer to Problem 47E

Sum of the geometric sequence is $\boxed{S_9=40353606}$

Explanation of Solution

Given information:

  ${\displaystyle \sum _{i=1}^{9}6{\left(7\right)}^{i-1}}$

Formula used:

Nth term formula of a GP series $a_n=a_1{r}^{n-1}$

Sum of $n$ term in geometric progression $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$

Find the first term and the common ratio.

  $\begin{array}{l}{a}_1=6{\left(7\right)}^{1-1}=6\\ a_1=6\end{array}$ Identify first term.

  $r=7$ Identify the common ratio.

Now finding the sum of geometric progression using $S_n=\frac{a_1\left(1-r^n\right)}{1-r}$ :

  $S_9=\frac{a_1\left(1-r^9\right)}{1-r}$ Write rule for $S_9$

Putting value of $a_1$ and $r$ in the rule of $S_9$ :

  $\begin{array}{l}{S}_9=\frac{6\left(1-{\left(7\right)}^9\right)}{1-7}\\ \\ S_9=\frac{6\left(1-40353607\right)}{-6}\\ \\ S_9=-1\left(-40353606\right)\\ \\ \boxed{S_9=40353606}\end{array}$