What is the sum of this geometric series? \sum_(k-1)^4 6^(k-1)

Answer 1

To find the sum of the geometric series \(\sum_{k=1}^{4} 6^{k-1}\), we can use the formula:

\[S = \frac{a(1-r^n)}{1-r}\]

Where:
- \(S\) is the sum of the series
- \(a\) is the first term of the series
- \(r\) is the common ratio of the series
- \(n\) is the number of terms in the series

In this case:
- \(a = 6^{1-1} = 6^0 = 1\) (since \(k\) starts from 1)
- \(r = 6\)
- \(n = 4\) (since we are summing from \(k=1\) to \(k=4\))

Plugging these values into the formula:

\[S = \frac{1(1-6^4)}{1-6} = \frac{1-1296}{-5} = \frac{-1295}{-5} = 259\]

Therefore, the sum of the geometric series \(\sum_{k=1}^{4} 6^{k-1}\) is 259.