Questions: Evaluate:
Sum from n=1 to 6 of 4(3)^(n-1)
S=[?]
Remember: for a geometric series, S=a(1-r^n)/(1-r)
Transcript text: Evaluate:
\[
\begin{array}{c}
\sum_{n=1}^{6} 4(3)^{n-1} \\
S=[?]
\end{array}
\]
Remember: for a geometric series, $S=\frac{a\left(1-r^{n}\right)}{1-r}$
Solution
Solution Steps
Step 1: Identify the Parameters
We are given a geometric series defined by the sum:
\[
S = \sum_{n=1}^{6} 4(3)^{n-1}
\]
Here, the first term \(a = 4\), the common ratio \(r = 3\), and the number of terms \(n = 6\).
Step 2: Apply the Geometric Series Formula
To find the sum \(S\), we use the formula for the sum of a geometric series:
\[
S = \frac{a(1 - r^n)}{1 - r}
\]
Substituting the identified values:
\[
S = \frac{4(1 - 3^6)}{1 - 3}
\]
Step 3: Calculate the Sum
Calculating \(3^6\):
\[
3^6 = 729
\]
Now substituting back into the formula:
\[
S = \frac{4(1 - 729)}{1 - 3} = \frac{4(-728)}{-2} = \frac{2912}{2} = 1456
\]