0
$\begingroup$

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?

$\endgroup$
3
  • 1
    $\begingroup$ Consider a general case. Block them into $4$ consecutive numbers if you call the first one $n$. $\endgroup$ – mathreadler Oct 28 '17 at 9:59
  • 2
    $\begingroup$ 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 Oct 28 '17 at 13:17
  • 1
    $\begingroup$ 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 Aug 19 '19 at 19:29
6
$\begingroup$

It's $$(1+5+...+49)+(2+6+...+50)-(3+7+...+51)-(4+8+...+52)+51+52=$$ $$=\frac{(2+12\cdot4)13}{2}+\frac{(4+12\cdot4)13}{2}-\frac{(6+12\cdot4)13}{2}-\frac{(8+12\cdot4)13}{2}+103=$$ $$=(1+2-3-4)\cdot13+103=51.$$

$\endgroup$
5
  • $\begingroup$ Thanks. But I don't get the idea behind this formula you just did. $\endgroup$ – dimwitt04 Oct 28 '17 at 10:42
  • 2
    $\begingroup$ @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 Oct 28 '17 at 10:49
  • $\begingroup$ Excuse my ignorance but where does 13 and 12 come from? Is it the number of terms? $\endgroup$ – dimwitt04 Oct 28 '17 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 Oct 28 '17 at 11:43
  • 2
    $\begingroup$ Can down-voter explain us why did you do it? $\endgroup$ – Michael Rozenberg Jul 9 '19 at 14:25
19
$\begingroup$

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

$\endgroup$
10
$\begingroup$

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 $$\sum_{k=1}^n(-1)^{\left\lfloor\frac{k-1}{2}\right\rfloor}\cdot k =\begin{cases} -n&\text{if $n\equiv 0\pmod{4}$}\\ 1&\text{if $n\equiv 1\pmod{4}$}\\ n+1&\text{if $n\equiv 2\pmod{4}$}\\ 0&\text{if $n\equiv 3\pmod{4}$}.\\ \end{cases}$$

$\endgroup$
2
  • 2
    $\begingroup$ (+1) I was just about to post an answer similar to the first part of this. Good generalization! $\endgroup$ – robjohn Oct 28 '17 at 12:49
  • 1
    $\begingroup$ @Holo Our generalizations are equivalent... $\endgroup$ – Robert Z Oct 28 '17 at 14:13
9
$\begingroup$

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?

$\endgroup$
1
  • 2
    $\begingroup$ And if it ended with - 51 the sum would be 0... $\endgroup$ – hkBst Oct 28 '17 at 12:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.