Answer
$\sum_{k=1}^{11} (-1)^{k+1}(\frac{2^k}{3^k})$
Work Step by Step
We can see that this is the sum of the $6$ fractions in which the first denominator is $2$, and gets multiplied by $3$ in each term and in which the first numerator is $3$, and gets multiplied by $3$ in each term. Also, the term changes sign each time. Hence the summation formula is: $\sum_{k=1}^{11} (-1)^{k+1}(\frac{2^k}{3^k})$
