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)

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, $S=\frac{a\left(1-r^{n}\right)}{1-r}$
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Solution

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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: \[ S = \frac{4(1 - 729)}{1 - 3} = \frac{4(-728)}{-2} = \frac{2912}{2} = 1456 \]

Final Answer

The sum of the series is \[ \boxed{S = 1456} \]

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