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February 23rd, 2017, 06:26 PM   #1
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Techniques for adding the numbers from n to n

Hey guys, I have some homework which I can't figure out how to do.

The question is simple: Find a formula that can find the sum of A to B while adding 1 to A in a specific amount of times. Like adding 1 to A and adding that ontop of the previous A in n times.

To give you an idea of what I mean, it's basically the same as going into Excel, adding 1, 2, 3, 4 and so on in a column like A, then marking all of those numbers and getting the sum of it.
What would the formula for that be?

Picture: https://gyazo.com/1a25ae38cc1b9270e1a2d1db271611ca

The numbers on the left Y axis = amount of times
The numbers on the right side "1, 2, 3, 4" is A + 1
The SUM = A + 1 * amount of times.

So basically, does anyone know the formula that excel uses to the "autosum" function?

Last edited by emilo0212; February 23rd, 2017 at 07:00 PM. Reason: wrong question
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February 23rd, 2017, 06:33 PM   #2
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$\displaystyle{\sum_{k=m}^n}~k = $

$\displaystyle{\sum_{k=0}^{n-m}}~(k+m) = $

$\displaystyle{\sum_{k=0}^{n-m}}~k+\displaystyle{\sum_{k=0}^{n-m}}~m = $

$\dfrac{(n-m)(n-m+1)}{2} + m(n-m+1)=$

$(n-m+1)\left(\dfrac{n-m}{2} + m\right) =$

$(n-m+1)\left(\dfrac{n+m}{2}\right)$
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February 23rd, 2017, 06:34 PM   #3
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Quote:
Originally Posted by romsek View Post
$\displaystyle{\sum_{k=m}^n}~k = $

$\displaystyle{\sum_{k=0}^{n-m}}~(k+m) = $

$\displaystyle{\sum_{k=0}^{n-m}}~k+\displaystyle{\sum_{k=0}^{n-m}}~m = $

$\dfrac{(n-m)(n-m+1)}{2} + m(n-m+1)=$

$(n-m+1)\left(\dfrac{n-m}{2} + m\right) =$

$(n-m+1)\left(\dfrac{n+m}{2}\right)$
Yea I just changed my question, cuz I had the wrong idea of the question. Does this one work with the new question?
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February 23rd, 2017, 06:43 PM   #4
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Quote:
Originally Posted by emilo0212 View Post
Yea I just changed my question, cuz I had the wrong idea of the question. Does this one work with the new question?
If I understand what you wrote then $m=A, ~n=B$
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February 23rd, 2017, 06:44 PM   #5
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Quote:
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If I understand what you wrote then $m=A, ~n=B$
I see, could you read the last bit I just updated and see if it still works?

and are you asking if m = A and n = B or are you saying that m would be A and n would be B
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February 23rd, 2017, 06:51 PM   #6
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To ensure we understand your requirement correctly, can you give a simple numerical example, complete with all the calculated results?
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February 23rd, 2017, 06:52 PM   #7
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Quote:
Originally Posted by emilo0212 View Post
I see, could you read the last bit I just updated and see if it still works?

and are you asking if m = A and n = B or are you saying that m would be A and n would be B
I'm saying that in the formula I gave you in post two, substitute $A$ for $m$ and $B$ for $n$
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February 23rd, 2017, 06:53 PM   #8
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Quote:
Originally Posted by skipjack View Post
To ensure we understand your requirement correctly, can you give a simple numerical example, complete with all the calculated results?
Yes of course, here you have it and I'll update my post quickly with it.

https://gyazo.com/1a25ae38cc1b9270e1a2d1db271611ca
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February 23rd, 2017, 06:54 PM   #9
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Quote:
Originally Posted by romsek View Post
I'm saying that in the formula I gave you in post two, substitute $A$ for $m$ and $B$ for $n$
Ahh, okay, well I'll try that out, although I dont think that's the case since I probably explained the question in a wrong way. I'll update my question now with a picture, so you can compare & understand.
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February 23rd, 2017, 07:10 PM   #10
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$A + (A+1) + (A+2) \dots + (B-1) + B = $

$\displaystyle{\sum_{k=A}^B}~k = (B-A+1)\left(\dfrac{A+B}{2}\right)$
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