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$$\sum_{k=0}^{2n+1} (-1)^kk $$

The answer given is $-n-1$. I have searched for how to do it, but I have problems simplifying the sum and solving it. How do you go about solving this?

  • 6
    Try it for a few particular cases like $n=1$, then $n=2$ and few more and see the pattern. – 2017-01-03

5 Answers 5

17

Group the terms together in pairs: $$\sum_{k=0}^{2n+1} (-1)^kk = \sum_{k=0}^{n} (2k - (2k+1)) = \sum_{k=0}^{n} (-1) = -(n+1)$$

16

Just write it out: \begin{aligned} \sum_{k=0}^{2n+1} (-1)^kk&=0-1+2-3+\cdots+2n-(2n+1)\\ &=(0-1)+(2-3)+\cdots+[2n-(2n+1)]\\ &=(-1)+(-1)+\cdots+(-1)\\ &=(n+1)(-1)\\ &=-n-1. \end{aligned}

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    "Just write it out" = induction? – 2017-01-03
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    @SimpleArt Perhaps but it's a simple problem so I think explicit visualization is more helpful than making it formal. – 2017-01-03
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    Just to point out to future users that these visual arguments are usually "hidden induction". And yes, simple problem, simple answer, very beautiful IMO. – 2017-01-03
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    @SimpleArt I don't think there's any need for induction here, hidden or not. The argument is simply that there are $2n+2$ terms $t_0, \dots, t_{2n+1}$ in the sum and, for each $i\in\{0, \dots, n+1\}$, we have $t_{2i}+t_{2i+1}=-1$, so the total is $(n+1)(-1)$. (This is just the same as [SteamyRoot's answer](http://math.stackexchange.com/a/2082212/97480), which was posted five seconds earlier.) – 2017-01-03
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    @DavidRicherby I would classify "dots" as hidden induction. At least, formalizing the proof above in any logical system I am aware of requires using some form of induction principle. – 2017-01-03
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    @chi OK but the practical, working mathematician isn't formalizing proofs in any form of logical system: they're writing a convincing argument. The arguments that I and SteamyRoot wrote involve no induction and SteamyRoot's doesn't even involve dots. If you want to reduce everything to some axiomatization then even the definition of the integers involves induction, so the statement "there's a hidden induction" becomes a tautology for almost any statement about integers. – 2017-01-03
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    For finite sums this works fine but you should be careful with this: the "just group into pairs of numbers" trick does not work, for example, for $\sum_{n = 0}^\infty (-1)^n$... – 2017-01-04
9

Split it into two series:

\begin{align*} \sum\limits_{k=0}^{2n+1}(-1)^kk &=\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}2k+1\right)\\ &=\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}2k\right)-\left(\sum\limits_{k=0}^{n}1\right)\\ &=-\sum\limits_{k=0}^{n}1\\ &=-(n+1)\,. \end{align*}

6

You have:

$$ \begin{align} \sum_{k=0}^{2n+1} (-1)^k k &= \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} (2k+1) \\ &= \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} 2k - \sum_{k=0}^{n} 1 \\ &= - \sum_{k=0}^{n} 1 \\ &= -(n+1) \end{align} $$

1

Another approach is to exploit the sum $$\sum_{k=0}^{2n+1}x^{k}=\frac{1-x^{2n+2}}{1-x}. $$ Taking the derivative we have $$\sum_{k=0}^{2n+1}kx^{k}=\sum_{k=1}^{2n+1}kx^{k}=x\frac{-\left(2n+2\right)x^{2n+1}\left(1-x\right)+1-x^{2n+2}}{\left(1-x\right)^{2}} $$ hence taking $x=-1 $ we get $$\sum_{k=0}^{2n+1}k\left(-1\right)^{k}=\color{red}{-\left(n+1\right)}$$ as wanted.