$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (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?

As mentioned by sharding4, simplify each parenthesized difference into $-1$ to achieve:
$$-(1+2+3+4+\dots+2016)+2017^2$$
Then recall the partial sum formula:
$$\sum_{k=1}^n k={n(n+1)\over 2}$$
Then apply:
$$-\left({2016\cdot2017\over 2}\right)+2017^2 = 2017k$$
Divide by $2017$:
$$\require{cancel}{-\left({2016\,\cdot\,2017\over 2}\right)+2017^2 \over 2017} = k$$
$${\cancelto{-1008}{{-\left({2016\,\cdot\,2017\over 2}\right) \over 2017}} + \cancelto{2017}{{2017^2\over 2017}}} = k$$
$$-1008+2017=k$$
$$k=1009$$
