Question

find the sum of the finite geometric series.\\(\sum_{n=1}^{6} 4 \cdot 3^{n - 1}\\)

Explanation:

Step1: Recall 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 \( S_n=\frac{a_1(1 - r^n)}{1 - r} \), where \( a_1 \) is the first term, \( r \) is the common ratio, and \( r
eq1 \). For a geometric series in the form \( \sum_{n = 1}^{n}a_1\cdot r^{n - 1}\), we can identify \( a_1 \) and \( r \) directly.

In the given series \( \sum_{n = 1}^{6}4\cdot3^{n - 1} \), we have \( a_1 = 4 \) (when \( n = 1 \), \( 4\cdot3^{1-1}=4\cdot1 = 4 \)) and \( r = 3 \), and \( n=6 \).

Step2: Substitute the values into the formula

Substitute \( a_1 = 4 \), \( r = 3 \), and \( n = 6 \) into the sum formula:

\( S_6=\frac{4(1 - 3^6)}{1 - 3} \)

First, calculate \( 3^6 \): \( 3^6=3\times3\times3\times3\times3\times3 = 729 \)

Then, calculate the numerator: \( 1-729=-728 \), so \( 4\times(1 - 729)=4\times(-728)=-2912 \)

Next, calculate the denominator: \( 1 - 3=-2 \)

Now, divide the numerator by the denominator: \( S_6=\frac{-2912}{-2}=1456 \)

Answer:

\( 1456 \)