## anonymous one year ago Find the indicated sum for the geometric series: ∑4k=1^3k−1

1. anonymous

2. Astrophysics

k = 1 is where you start, and 4 is where you end, so the following$\large \sum_{k=1}^{4} 3^{k-1} = 3^{1-1}+3^{2-1}+3^{3-1}+3^{4-1} = 3^0+3^1+3^2+3^3$

3. Astrophysics

Note your exponent rules here, $x^0 = 1$