# Sum of n consecutive numbers [duplicate]

Possible Duplicate:
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.): $10 + 11 + 12 + 13 = 46$.

• 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...... Jul 9, 2011 at 9:45
• Alternately, if there are $n$ successive integers, starting with $m+1$, the answer is $mn + (1 + 2 + \ldots + n)$. Jul 9, 2011 at 10:02
• I'm pretty sure this is a duplicate. :) Jul 9, 2011 at 10:15
• BTW: MarkUp is not allowed in subject title of posts. I've removed the asterisks. Jul 9, 2011 at 11:15
• This is not at all a duplicate of the question its marked as a dupe of. This is asking for a simple formula, the other a detailed proof. Very different things. StackExchange has an epidemic of busy bodies marking things as duplicates unnecessarily and destructively.
– B T
Dec 18, 2015 at 8:10

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

• ... and this works for any arithmetic progression Jul 9, 2011 at 10:20
• computation Sum = total_numbers * ( first / 2 + last / 2 ) Mar 14, 2017 at 10:53
• @Miguel I believe your order of operations is wrong. It should be: Sum = total_numbers * (first + last) / 2 Feb 28, 2021 at 2:48
• @PaulHazen well depends on who is the user, computers do the first/2 and add last/2 in this order, calculators might use the "human" order, which would end up in a total different result. Either way, yes a/2 + b/2 is (a+b)/2 so your solution is more clear and its shorter and removed doubts. Thanks. Mar 1, 2021 at 22:14

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: • Thanks Leon. Great illustration of the math.
– Carl
Jul 13, 2011 at 11:44