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Proof for formula for sum of sequence $1+2+3+\ldots+n$?

Is there a shortcut method to working out the sum of n consecutive positive integers?

Firstly, starting at $1 ... 1 + 2 + 3 + 4 + 5 = 15.$

Secondly, starting at any other positive integer ...($10$ e.g.): $10 + 11 + 12 + 13 = 46$.

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marked as duplicate by robjohn Nov 11 '12 at 12:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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The solution to the second problem follows from a solution of the first since $(m+1)+\cdots+n=S(n)-S(m)$ where $S(k)=1+\cdots+k$. About the first... the legend says that Gauss as a schoolboy realized quickly that if you sum as $(1+n)+(2+(n-1))+\ldots$ all the summands are the same, thus...... –  Andrea Mori Jul 9 '11 at 9:45
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Alternately, if there are $n$ successive integers, starting with $m+1$, the answer is $mn + (1 + 2 + \ldots + n)$. –  Geoff Robinson Jul 9 '11 at 10:02
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I'm pretty sure this is a duplicate. :) –  Beni Bogosel Jul 9 '11 at 10:15
    
BTW: MarkUp is not allowed in subject title of posts. I've removed the asterisks. –  Willie Wong Jul 9 '11 at 11:15
    
So I thought this was certainly a duplicate. But then I attempted to find such a duplicate, but to no avail. I'm at a loss as to how to effectively search for such a duplicate now. –  mixedmath Jul 9 '11 at 15:49
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2 Answers

up vote 14 down vote accepted

Take the average of the first number and the last number, and multiply by the number of numbers.

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... and this works for any arithmetic progression –  Henry Jul 9 '11 at 10:20
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The rule, as given by Gerry's answer (and the generalization as per Henry's comment) can be easily visualized, in a similar way as we deduce the area of a rectangular trapezium:

enter image description here

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Thanks Leon. Great illustration of the math. –  Carl Jul 13 '11 at 11:44
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