QuestionFind the sum of the given finite geometric series. $6+\frac{6}{3}+\frac{6}{9}+\frac{6}{27}+\ldots+\frac{6}{59049}$
The sum of the finite geometric series is $\square$ (Type an integer or a simplified fraction.)

Studdy Solution

STEP 1

1. The series is a finite geometric series.
2. We need to identify the first term, the common ratio, and the number of terms.
3. Use the formula for the sum of a finite geometric series to find the sum.

STEP 2

1. Identify the first term, common ratio, and number of terms in the series.
2. Use the formula for the sum of a finite geometric series.
3. Calculate the sum.

STEP 3

Identify the first term $a$ , the common ratio $r$ , and the number of terms $n$ in the series.
The first term $a$ is the first term of the series: $a = 6$
The common ratio $r$ is the factor by which each term is multiplied to get the next term: $r = \frac{6/3}{6} = \frac{1}{3}$
To find the number of terms $n$ , observe the last term of the series: $\frac{6}{59049} = 6 \times \left(\frac{1}{3}\right)^{n-1}$
Since $59049 = 3^{10}$ , we have: $\left(\frac{1}{3}\right)^{n-1} = \frac{1}{3^{10}}$ $n-1 = 10$ $n = 11$

STEP 4

Use the formula for the sum of a finite geometric series:
The formula for the sum $S_n$ of the first $n$ terms of a geometric series is: $S_n = a \frac{1 - r^n}{1 - r}$
Substitute the identified values into the formula: $S_{11} = 6 \frac{1 - \left(\frac{1}{3}\right)^{11}}{1 - \frac{1}{3}}$

STEP 5

Calculate the sum:
First, calculate $\left(\frac{1}{3}\right)^{11}$ : $\left(\frac{1}{3}\right)^{11} = \frac{1}{177147}$
Substitute back into the sum formula: $S_{11} = 6 \frac{1 - \frac{1}{177147}}{\frac{2}{3}}$
Simplify the expression: $S_{11} = 6 \times \frac{177146}{177147} \times \frac{3}{2}$
$S_{11} = 9 \times \frac{177146}{177147}$
Since $\frac{177146}{177147}$ is very close to 1, the sum is approximately: $S_{11} = 9$
The sum of the finite geometric series is:
$\boxed{9}$

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