Questions: sum from n=1 to 6 of 3(4)^(n-1) S=[?]

Transcript text: \[ \sum_{n=1}^{6} 3(4)^{n-1} \\ S=[?] \]

Solution

Solution Steps

To evaluate the given sum, we recognize it as a geometric series. The general form of a geometric series is \( \sum_{n=0}^{N-1} ar^n \), where \( a \) is the first term and \( r \) is the common ratio. In this case, the series starts at \( n=1 \), so we adjust the formula accordingly. We can use the formula for the sum of the first \( N \) terms of a geometric series: \( S_N = a \frac{r^N - 1}{r - 1} \).

Solution Approach

Identify the first term \( a = 3 \) and the common ratio \( r = 4 \).
Use the formula for the sum of the first \( N \) terms of a geometric series.
Substitute \( a \), \( r \), and \( N = 6 \) into the formula to find the sum.

Step 1: Identify the Series

We are given the series \( S = \sum_{n=1}^{6} 3(4)^{n-1} \). This is a geometric series where the first term \( a = 3 \) and the common ratio \( r = 4 \).

Step 2: Apply the Geometric Series Formula

The sum of the first \( N \) terms of a geometric series can be calculated using the formula: \[ S_N = a \frac{r^N - 1}{r - 1} \] Substituting the values \( a = 3 \), \( r = 4 \), and \( N = 6 \) into the formula gives: \[ S_6 = 3 \frac{4^6 - 1}{4 - 1} \]

Step 3: Calculate the Sum

Calculating \( 4^6 \): \[ 4^6 = 4096 \] Now substituting back into the formula: \[ S_6 = 3 \frac{4096 - 1}{3} = 3 \frac{4095}{3} = 4095 \]