Answer
$1-\displaystyle \frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots+(-1)^{6}\left(\frac{1}{3^{6}}\right)=\sum_{k=0}^{6}(-1)^{k}\left(\frac{1}{3^{k}}\right)$
Work Step by Step
There are $7$ terms (terms contain from $\displaystyle \frac{1}{3^{0}}$ to $\displaystyle \frac{1}{3^{6}}$).
Count them with index k: $\displaystyle \sum_{k=0}^{6}$(...)
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=$0$) is $1=\displaystyle \frac{1}{3^{0}}$,
The second term (when k=$1$) is $-\displaystyle \frac{1}{3}=(-1)\frac{1}{3^{1}}$,
The third term (when k=$2$) is $+\displaystyle \frac{1}{9}=(-1)^{2}\frac{1}{3^{2}}$,
The kth is $(-1)^{k}\displaystyle \left(\frac{1}{3^{k}}\right)$ so
$1-\displaystyle \frac{1}{3}+\frac{1}{9}-\frac{1}{27}+\dots+(-1)^{6}\left(\frac{1}{3^{6}}\right)=\sum_{k=0}^{6}(-1)^{k}\left(\frac{1}{3^{k}}\right)$
