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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: 2,530 Thanks: 1390 $\displaystyle{\sum_{k=m}^n}~k =$ $\displaystyle{\sum_{k=0}^{n-m}}~(k+m) =$ $\displaystyle{\sum_{k=0}^{n-m}}~k+\displaystyle{\sum_{k=0}^{n-m}}~m =$ $\dfrac{(n-m)(n-m+1)}{2} + m(n-m+1)=$ $(n-m+1)\left(\dfrac{n-m}{2} + m\right) =$ $(n-m+1)\left(\dfrac{n+m}{2}\right)$ 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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 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: 20,926 Thanks: 2205 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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 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: 2,530 Thanks: 1390 $A + (A+1) + (A+2) \dots + (B-1) + B =$ $\displaystyle{\sum_{k=A}^B}~k = (B-A+1)\left(\dfrac{A+B}{2}\right)$ Tags adding, numbers, techniques Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post thomas2608 Calculus 2 November 19th, 2014 12:18 PM mathnoob2 Algebra 2 May 31st, 2013 04:53 AM TabbiSue Advanced Statistics 2 October 23rd, 2012 07:52 AM grogmachine Computer Science 10 June 20th, 2011 05:05 AM MRKhal Algebra 2 April 2nd, 2011 01:17 AM

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