Answer
$S_n=1-2^n$
Work Step by Step
The sum of the first $n$ terms of a Geometric Sequence is given by:
$S_{n}=\displaystyle\sum_{k=1}^n a_1r^{k-1}=a_{1} (\dfrac{1-r^{n}}{1-r}) ; \ r\neq 0,1$
We are given:
$a_{1}= -1 ; \ r= 2$
Now, $S_n= (-1) \ [\dfrac{1-(2)^{n}}{1- 2} \ ] \\= -1 \ [\dfrac{1-(2)^{n}}{-1} \ ] $
Therefore, $S_n=1-2^n$