Yes, the proof is now taught in elementary school, as I recall. It was the
first one I ever learned to do.
It clearly does work, with adjustment, for ranges starting with other than 1,
and it works with negative numbers too. It also doesn't matter whether you are
counting down or counting up, because it is symmetrical as far as addition is
concerned.
To go from m to n, assume m is not larger than n. If not, simply interchange
them since the summation taken in either direction is the same.
Write down the numbers
m, m+1, ..., n-1, n
n, n-1, ..., m+1, m
Each of those rows lists the numbers to be added, but in reversed sequence, as
Rob shows, below.
Notice that each vertical pair adds up to precisely m+n.
Satisfy yourself that there are exactly n-m+1 pairs (the columns) like that.
Notice that works for m=n too.
So the sum of all the numbers (using both rows) is (n-m+1)*(m+n)
But that's double the sum of just one of the rows. The sum of just one of
those rows, either one,
is (n-m+1)*(m+n)/2. (The sum over both rows must be an even number so the
division always work.)
Notice that m=0 and m=1 fall out as having the same results.
- Dennis
PS: That's why there is no need for a special function in OpenFormula to
accomplish this. Gauss showed that it is not hard. I once owned a book that
had many simplified formulas for series and summations. It's probably still
available from Dover Publications. Or just use Internet Search and find
Wikipedia articles and YouTube videos about them.
PPS: In elementary school, our principal gave out a problem based on the story
of the fellow who was to be rewarded by a king and had asked for grains of
wheat to be put on the squares of a chess board (64 squares) with 1 on the
first square, 2 on the second, 4 on the third, etc., until the end. In the
story, the fellow was rewarded by beheading. Our school principal wanted to
know how may grains of wheat would have been awarded. Every software developer
should know the shortcut, that requires only doublings and one subtraction.
You never have to add up all of the individual numbers, so the chance for error
is significantly reduced. I didn't figure that out, but the principal showed
me when I took in my clumsy column of numbers. At that time, commercial
computers were just coming out, so binary arithmetic was not widely-known.
-----Original Message-----
From: Rob Weir [mailto:[email protected]]
Sent: Saturday, June 22, 2013 07:54 AM
To: [email protected]
Subject: Re: Calc: Easy way to do N+(N-1)+(N-2)+(N-3)..(N-N+1)
On Fri, Jun 21, 2013 at 6:39 PM, Toki Kantoor <[email protected]> wrote:
> On 06/21/2013 06:37 PM, Brian Barker wrote:
>> Yes. Sigma (1 to n) is n(n+1)/2.
>
> Thanks.
>
>>> If so, what is the formula, extension, or something?
>>
>> =Xn*(Xn+1)/2
>>
>> I trust this helps.
>
> It helps a lot.
There is a famous story about this formula. The mathematician Carl
Friedrich Gauss, when he was 10 years old, was in an arithmetic class
where the teacher gave them all the problem to sum the digits from 1
to 100.
Maybe the teacher had some other task he wanted to do, or wanted to
take a nap? So he gave them this task to keep them busy.
Gauss figured this out in his head (the answer is 5050), by
discovering the above formula, and put down his slate, much angering
the teacher.
The key is to rearrange the calculation. So instead of
1+2+3+4...+100, think of it as: (1+100) + (2+99) + (3+98) ... +
(49+52) + (50+51) = 101*50
Regards,
-Rob
> I wish I had book that contained functions and their formula, that
> somebody that never took a match course could understand.
>
> jonathon
> --
> LibreOffice in a Multi-Lingual Environment.
>
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