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 February 23rd, 2017, 06:26 PM #1 Newbie   Joined: Feb 2017 From: Denmark Posts: 12 Thanks: 1 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
 February 23rd, 2017, 06:33 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 $\displaystyle{\sum_{k=m}^n}~k =$ $\displaystyle{\sum_{k=0}^{n-m}}~(k+m) =$ $\displaystyle{\sum_{k=0}^{n-m}}~k+\displaystyle{\sum_{k=0}^{n-m}}~m =$ $\dfrac{(n-m)(n-m+1)}{2} + m(n-m+1)=$ $(n-m+1)\left(\dfrac{n-m}{2} + m\right) =$ $(n-m+1)\left(\dfrac{n+m}{2}\right)$
February 23rd, 2017, 06:34 PM   #3
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 Originally Posted by romsek $\displaystyle{\sum_{k=m}^n}~k =$ $\displaystyle{\sum_{k=0}^{n-m}}~(k+m) =$ $\displaystyle{\sum_{k=0}^{n-m}}~k+\displaystyle{\sum_{k=0}^{n-m}}~m =$ $\dfrac{(n-m)(n-m+1)}{2} + m(n-m+1)=$ $(n-m+1)\left(\dfrac{n-m}{2} + m\right) =$ $(n-m+1)\left(\dfrac{n+m}{2}\right)$
Yea I just changed my question, cuz I had the wrong idea of the question. Does this one work with the new question?

February 23rd, 2017, 06:43 PM   #4
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Quote:
 Originally Posted by emilo0212 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 $m=A, ~n=B$

February 23rd, 2017, 06:44 PM   #5
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Quote:
 Originally Posted by romsek 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

 February 23rd, 2017, 06:51 PM #6 Global Moderator   Joined: Dec 2006 Posts: 18,954 Thanks: 1601 To ensure we understand your requirement correctly, can you give a simple numerical example, complete with all the calculated results?
February 23rd, 2017, 06:52 PM   #7
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Quote:
 Originally Posted by emilo0212 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 $n$

February 23rd, 2017, 06:53 PM   #8
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 Originally Posted by skipjack 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

February 23rd, 2017, 06:54 PM   #9
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 Originally Posted by romsek 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.

 February 23rd, 2017, 07:10 PM #10 Senior Member     Joined: Sep 2015 From: USA Posts: 1,931 Thanks: 999 $A + (A+1) + (A+2) \dots + (B-1) + B =$ $\displaystyle{\sum_{k=A}^B}~k = (B-A+1)\left(\dfrac{A+B}{2}\right)$

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