Answer
$S_{1}=-1$
$S_{2}=0$
$S_{3}=-1$
$S_{4}=0$
$S_{5}=-1$
$S_{6}=0$
Work Step by Step
We are given:
$-1, 1, -1, 1$ ...
We notice that the terms alternate between $-1$ and $1$:
$a_{1}=-1=(-1)^1$
$a_{2}=1=(-1)^2$
$a_{3}=-1=(-1)^3$
$a_{4}=1=(-1)^4$
Therefore:
$a_{n}=(-1)^{n}$
So:
$a_{5}=-1$
$a_{6}=1$
We find the partial sums:
$S_{1}=-1$
$S_{2}=-1+1=0$
$S_{3}=0-1=-1$
$S_{4}=-1+1=0$
$S_{5}=0-1=-1$
$S_{6}=-1+1=0$