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Today I tried to compute $$ \sum_{k =1}^{100}(-1)^k $$

Is there a way to find the result more quickly ?

Below if my attempt to find the result.

Especially without considering the case of odd and even numbers like I did ?


Let's consider the following sum:

$$ \sum_{k=1}^{n}(-1)^k = (-1)^1 + (-1)^2 + (-1)^3 + ... +(-1)^n $$

If $n$ is even then $\frac n2 $ terms have an even exponent and $\frac n2 $ terms have an odd exponent.

$$(-1)^p = -1 \tag{when $p$ is odd} $$ $$ (-1)^p = 1 \tag{when $p$ is even} $$

Then when $n$ is even

$$\begin{align} \sum_{k=1}^{n}(-1)^k &= \frac n2 \times(-1) + \frac n2 \times 1 \\ & = \frac n2 - \frac n2 \\ & = 0 \\ \end{align}$$

100 is an even number, so

$$ \sum_{k =1}^{100}(-1)^k = 0 $$

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    $\begingroup$ Call the sum $S$, so that $S=(-1)+(1)+\dotsb+(-1)+(1)$. Notice that $-S=(1)+(-1)+\dotsb+(1)+(-1)$—that is, $-S$ is just $S$ reversed. So, $S=-S$, which means that $S=0$. $\endgroup$ – Akiva Weinberger Sep 23 '14 at 20:15
  • $\begingroup$ If you say that $n/2$ terms have an even exponent, then you are already assuming that $n$ is even. $\endgroup$ – Dietrich Burde Sep 23 '14 at 20:15
  • $\begingroup$ The quick way is to cancel $1$s with $-1$s. You have equally many of both, and if you didn't, you'd have just one term left after canceling. But of course it's also a finite geometric series, and there's a standard formula for that. $\endgroup$ – Michael Hardy Sep 23 '14 at 20:57
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yes the series has a direct summation formula: $$ S_k=\sum_{i=1}^k (-1)^i={1 \over 2}\Big((-1)^k-1\Big) $$ you can prove this easily using induction. $$\begin{aligned} \mbox{k=1: }\ \ \ &true\\ \mbox{k }\to\mbox{ k+1: }\ \ \ &S_{k+1}=S_{k}+(-1)^{k+1}&={1 \over 2}\Big((-1)^k-1\Big)+(-1)^{k+1} =\\ & &={1 \over 2}\Big((-1)^k+2(-1)^{k+1}-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1+2(-1)^{1})-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(1-2)-1\Big) =\\ & &={1 \over 2}\Big((-1)^k(-1)-1\Big) =\\ & &={1 \over 2}\Big((-1)^{k+1}-1\Big) \ \ \ \square \end{aligned} $$

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Note that $(-1)^{2k-1}=-1$ and $(-1)^{2k}=1$ for all $k\geq1$. Hence the sum upto any even power should equal $0$.

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The easiest way to find the result is to look at the first few partial sums

$$ -1, 0, -1, \ldots $$

and identify the pattern. Everything beyond that is just seeking to give a more rigorous justification of it.

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  • $\begingroup$ Sure. I think that's what I am looking for. A rigorous justification but explained quickly. $\endgroup$ – alexandrekow Sep 23 '14 at 20:37
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Since $(-1)^{2n}=1$ and $(-1)^{2n-1}=-1$, we have $$\sum_{k=1}^n (-1)^k = -1+1-1+1-\cdots \pm 1=\left\{\begin{array}{ll}-1 & \text{if}\,n\,\text{is odd}\\ 0 & \text{if}\,n\,\text{is even}\end{array}\right.$$

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$$\sum_{k=1}^{n}(-1)^k=\sum_{k=1}^{n}(-1)(-1)^{k-1}=-\sum_{k=1}^{n}(-1)^{k-1}=$$ $$=-\sum_{k=0}^{n-1}(-1)^{k}=-\frac{1-(-1)^{n}}{1-(-1)}=\frac{(-1)^{n}-1}{2}$$ for $n=100$ $$\sum_{k=1}^{100}(-1)^k=\frac{(-1)^{100}-1}{2}=\frac{1-1}{2}=0$$

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  • $\begingroup$ How do you do to transform the third part of the first line into the fourth one? $\endgroup$ – alexandrekow Sep 23 '14 at 20:39
  • $\begingroup$ This is of course the hard way. $\endgroup$ – Michael Hardy Sep 23 '14 at 20:57
  • $\begingroup$ you can see my edit! $\endgroup$ – Adi Dani Sep 23 '14 at 20:59

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