Answer
$\sum_{k=0}^6\ (-1)^{k}\frac{1}{3^{k}}$
Work Step by Step
Based on the given summation, we notice that the nth term contains a sign value $(-1)^{n-1}$ and a power $\frac{1}{3^{n-1}}$, thus we have the sum as $\sum_{k=1}^n\ (-1)^{k-1}\frac{1}{3^{k-1}}$ with $n=7$ to get $\sum_{k=1}^7\ (-1)^{k-1}\frac{1}{3^{k-1}}$ or if we allow the starting term as $k=0$ to get $\sum_{k=0}^6\ (-1)^{k}\frac{1}{3^{k}}$
