## Precalculus (10th Edition)

$\sum_{k=0}^{6} (-1)^k(\frac{1}{3^k})$
We can see that this is the sum of the $6$ fractions in which the denominator is $1$ and in which the first numerator is $1$, and gets multiplied by $3$ in each term. Also, the term changes sign each time. Hence the summation formula is: $\sum_{k=0}^{6} (-1)^k(\frac{1}{3^k})$