$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (Solve for $k$) Question:

$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ 
Solve for $k$

My attempt:
$$1^2-2^2+3^2-4^2+…-2016^2+2017^2\\
\begin{align}= (1-2)(1+2)+(3-4)(3+4)+…+(2015-2016)(2015+2016)+2017^2 \end{align}
$$
What should I do next?

A very simple way of seeing this is to write the following:
$$\begin{align} \\
1^2-2^2+\dots-2016^2+2017^2 &= 1+(3^2-2^2)+\dots+(2017^2-2016^2) \\
&=1+(3-2)(3+2)+(5-4)(5+4)+\dots+(2017-2016)(2016+2017) \\
&=1+2+3+...+2016+2017 \\
&=\frac{(2018)(2017)}{2}=2017k \\
\end{align}$$
Which means that $k=\frac{2018}{2}=1009$.
