Answer
$\displaystyle \frac{2}{3}-\frac{4}{9}+\frac{8}{27}-\dots+(-1)^{11+1}\left(\frac{2}{3}\right)^{11}=\sum_{k=1}^{11}(-1)^{k+1}\left(\frac{2}{3}\right)^{k}$
Work Step by Step
There are 11 terms ( terms contain from $(\displaystyle \frac{2}{3})^{1}$ to $(\displaystyle \frac{2}{3})^{11}$.
Count them with index k: $\displaystyle \sum_{k=1}^{11}$(...)
Terms alternate in signs, so they will contain a power of (-1).
When the exponent $(-1)^{n}$ is odd, we have $-1$,
and when it is even, $+1.$
The first term (when k=$1$) is $+\displaystyle \frac{2}{3}=(-1)^{1+1}\left(\frac{2}{3}\right)^{1}$,
The second term (when k=$2$) is $-\displaystyle \frac{4}{9}=(-1)^{2+1}\left(\frac{2}{3}\right)^{2}$,
The third term (when k=$3$) is $+\displaystyle \frac{8}{27}=(-1)^{3+1}\left(\frac{2}{3}\right)^{3}$,
The kth is $(-1)^{k+1}\displaystyle \left(\frac{2}{3}\right)^{k}$ so
$\displaystyle \frac{2}{3}-\frac{4}{9}+\frac{8}{27}-\dots+(-1)^{11+1}\left(\frac{2}{3}\right)^{11}=\sum_{k=1}^{11}(-1)^{k+1}\left(\frac{2}{3}\right)^{k}$
