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

Question

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?

Answer 1

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 $2017$:

$$\require{cancel}{-\left({2016\,\cdot\,2017\over 2}\right)+2017^2 \over 2017} = k$$ $${\cancelto{-1008}{{-\left({2016\,\cdot\,2017\over 2}\right) \over 2017}} + \cancelto{2017}{{2017^2\over 2017}}} = k$$ $$-1008+2017=k$$ $$k=1009$$

Answer 2

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 $k=\frac{2018}{2}=1009$.