Question: What is the sum of this geometric series? \[ \sum_{k=1}^{5} 6(4)^{k-1} \] 1, 536 8, 190 2,…

What is the sum of this geometric series?

\[ \sum_{k=1}^{5} 6(4)^{k-1} \]

1, 536

8, 190

2, 046

4, 096

Solution

The problem asks for the sum of the geometric series: \[ \sum_{k=1}^{5} 6(4)^{k-1} \] Identify the first term \(a\) and the common ratio \(r\). The first term \(a = 6(4)^{1-1} = 6 \cdot 1 = 6\). The common ratio \(r = 4\). The number of terms \(n = 5\). The formula for the sum \(S_n\) of the first \(n\) terms of a geometric series is: \[ S_n = a \frac{r^n - 1}{r - 1} \] Substitute the known values into the formula: \[ S_5 = 6 \frac{4^5 - 1}{4 - 1} \] Calculate \(4^5\): \[ 4^5 = 1024 \] Substitute back into the formula: \[ S_5 = 6 \frac{1024 - 1}{3} \] Simplify the expression: \[ S_5 = 6 \frac{1023}{3} \] Compute \(\frac{1023}{3}\): \[ \frac{1023}{3} = 341 \] Finally, calculate the sum: \[ S_5 = 6 \cdot 341 = 2046 \] The sum of the geometric series is 2,046.