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How does one evaluate the sum $1+2-3-4+5+6-7-8+\cdots+50$?

I know how to find the sum of arithmetic progressions: without the negative signs, one simply has $$ 1+2+\cdots+50=\frac{1}{2}\cdot(1+50)\cdot 50=51\times 25=1275. $$

But how does one calculate the one above?

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    $\begingroup$ Consider a general case. Block them into $4$ consecutive numbers if you call the first one $n$. $\endgroup$
    – mathreadler
    Commented Oct 28, 2017 at 9:59
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    $\begingroup$ Using @Robertz 's idea: better $1+2-3=0$, $-4+5+6-7=0$, $-8+9+10-11=0$,... $...-47=0$, $-48+49+50-51=0$ then correct for the last sum $0+51=51$ $\endgroup$
    – Gottfried Helms
    Commented Oct 28, 2017 at 13:17
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    $\begingroup$ I find it disheartening that this question has received so many downvotes, and has been closed and deleted, and noone has commented as to why...... $\endgroup$
    – user1729
    Commented Aug 19, 2019 at 19:29

4 Answers 4

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Look at the following: $$1+\overbrace{(2-3)}^{-1}+\overbrace{(-4+5)}^1+\cdots+50$$

So you have $1+\overbrace{-1+1\cdots}^{\frac{48}2=24\text{ times}}+50$ and because $24$ is even the middle part become $0$ and you left with $1+50=51$ and done

moreover, you can generalize it:$$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\times k=\begin{cases}n+1 &\text{if}\,\,\,(-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv0\pmod{2}\\ 1 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=1,&n\equiv1\pmod{2}\\ -n &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv0\pmod{2}\\ 0 &\text{if}\,\,\, (-1)^{\left\lfloor\frac{n-1}{2}\right\rfloor}=-1,&n\equiv1\pmod{2} \end{cases} $$

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Note that your sum can be written as $$\underbrace{[(1-3)+(2-4)]}_{-4}+\underbrace{[(5−7)+(6−8)]}_{-4}+\dots +\underbrace{[(45−47)+(46−48)]}_{-4}+49+50$$ that is $-4\cdot(48/4)+49+50=-48+49+50=51.$ More generally $$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\cdot k =\begin{cases} -n&\text{if $n\equiv 0\pmod{4}$}\\ 1&\text{if $n\equiv 1\pmod{4}$}\\ n+1&\text{if $n\equiv 2\pmod{4}$}\\ 0&\text{if $n\equiv 3\pmod{4}$}.\\ \end{cases}$$

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    $\begingroup$ (+1) I was just about to post an answer similar to the first part of this. Good generalization! $\endgroup$
    – robjohn ♦
    Commented Oct 28, 2017 at 12:49
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    $\begingroup$ @Holo Our generalizations are equivalent... $\endgroup$
    – Robert Z
    Commented Oct 28, 2017 at 14:13
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Note that $1+2-3=0$. Moreover, you will have -4+5+6-7 and so on... if you consider pairs of numbers, you will always have +1. How much times do you do this computation?

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    $\begingroup$ And if it ended with - 51 the sum would be 0... $\endgroup$
    – hkBst
    Commented Oct 28, 2017 at 12:00
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It's $$(1+5+...+49)+(2+6+...+50)-(3+7+...+51)-(4+8+...+52)+51+52=$$ $$=\frac{(2+12\cdot4)13}{2}+\frac{(4+12\cdot4)13}{2}-\frac{(6+12\cdot4)13}{2}-\frac{(8+12\cdot4)13}{2}+103=$$ $$=(1+2-3-4)\cdot13+103=51.$$

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  • $\begingroup$ Thanks. But I don't get the idea behind this formula you just did. $\endgroup$
    – dimwitt04
    Commented Oct 28, 2017 at 10:42
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    $\begingroup$ @dimwitt04 It's a formula for the sum of the arithmetic progression: $S_{n}=\frac{(2a_1+(n-1)d)n}{2}$. $\endgroup$
    – Michael Rozenberg
    Commented Oct 28, 2017 at 10:49
  • $\begingroup$ Excuse my ignorance but where does 13 and 12 come from? Is it the number of terms? $\endgroup$
    – dimwitt04
    Commented Oct 28, 2017 at 11:41
  • $\begingroup$ @dimwitt04 Yes of course! If $n$ is a number of terms we obtain: $49=1+(n-1)4$, which gives $n=13$. I used $a_n=a_1+(n-1)d$. $\endgroup$
    – Michael Rozenberg
    Commented Oct 28, 2017 at 11:43
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    $\begingroup$ Can down-voter explain us why did you do it? $\endgroup$
    – Michael Rozenberg
    Commented Jul 9, 2019 at 14:25

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