Question

find the sum of the finite geometric series.
\\(\sum_{n=1}^{6} 2 \cdot 4^{n}\\)

Explanation:

Step1: Recall the formula for the sum of a finite geometric series

The sum of a finite geometric series \(\sum_{n = 1}^{k} a\cdot r^{n}\) is given by \(S=\frac{a\cdot r(r^{k}-1)}{r - 1}\), where \(a\) is the coefficient, \(r\) is the common ratio, and \(k\) is the number of terms. In our series \(\sum_{n = 1}^{6}2\cdot4^{n}\), we have \(a = 2\), \(r=4\), and \(k = 6\).

Step2: Substitute the values into the formula

Substitute \(a = 2\), \(r = 4\), and \(k=6\) into the formula \(S=\frac{a\cdot r(r^{k}-1)}{r - 1}\).
First, calculate \(r^{k}=4^{6}\). We know that \(4^{6}=4096\).
Then, \(r^{k}-1=4096 - 1=4095\).
Next, \(a\cdot r=2\times4 = 8\).
Now, \(S=\frac{8\times4095}{4 - 1}\).
Calculate the numerator: \(8\times4095 = 32760\).
Calculate the denominator: \(4-1 = 3\).
Then, \(S=\frac{32760}{3}=10920\).

Answer:

\(10920\)