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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?

student
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    Can you simplify $(1-2), (3-4), \dots ,(2015-2016)$? – sharding4 Mar 10 '18 at 03:43
  • @sharding4 Yes I can, It's $-(1+2+3+4+…+2016)+2017^2$ – student Mar 10 '18 at 03:47
  • This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. The number 2017, in particular, makes this look like a homework or contest problem, so please provide a specific source. – Carl Mummert Mar 10 '18 at 03:52
  • Now can you sum the parenthesized expression? – NickD Mar 10 '18 at 03:54
  • Well, try something. 1-4=-3;-3+9=6;6-16=-10;-10+25=15$. See any pattern. And if $n^2=1+3+5+7+.....+(2n+1)$ does that pattern make some sort of sense? – fleablood Mar 10 '18 at 08:07

2 Answers2

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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 $2017$:

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

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

Heatconomics
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