Question

Find the indicated sum for the geometric series:

∑4k=1^3k−1

Answer 1

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

Answer 2

Note your exponent rules here, \[x^0 = 1\]