Answer
$\sum\limits_{n=1}^6(-1)^{n+1}n$
Work Step by Step
We have to rewrite the sum $1-2+3-4+5-6$ using sigma notation.
At first we have to find the general term. We have:
$a_1=1$
$a_2=-2$
$a_3=3$
$a_4=-4$
.....
It is easy to see that the even terms are $-n$, while the odd terms are $n$. So we can write:
$a_n=(-1)^{n+1}n$.
Now we count the number of terms in the sum: $6$, therefore the index will go from $1$ to $6$.
$$1-2+3-4+5-6=\sum\limits_{n=1}^6(-1)^{n+1}n$$
