Is there a formula for adding sequences of numbers, say the sum of numbers 1-25?

I've always just added the first and last numbers, then multiplied the result by half of the last number.

1+25=26
26*(25/2)=325

Is this the simplest solution?

If I remember correctly, one way to do it is

N(½N) + ½N

In the series 1,2,3...10, that would be

10(5) + 5

55

In the series 1,2,3...100, it would be

100(50) + 50

5050


The last one was a question in last Saturday's Trivia contest. Yes, we got it right, and we won easily.

The above formula may not work for a series with an odd number of components. I'm pretty sure there is a different formula for series ending in odd numbers.


The way to understand the development of the formula is to see that in any series with an even number of components, you will have pairs of numbers adding up to the final number.

Using 1 - 10, we have
10 + 0
9 + 1
8 + 2
7 + 3
6 + 4

and then we have half of the last number left unused, and thus needing to be added to the total.

They key to remembering this way tofigure it out when you have forgotten the formula is to remember the zero. Add it to the last number, and the rest fall into place.
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Sorry if I didn't explain this well. I am better explaining in person, where I can wave my arms around. It does help.

As they say, ∑n = n(n+1)/2.

And it works for even or odd n.

This is essentially the same formula that DG gave, though looked at a little differently. There is a famous story involving the great mathematician Carl Friedrich Gauss (1777-1855) that

quote:

...has it that in primary school his teacher, J.G. Büttner, tried to occupy pupils by making them add up the integers from 1 to 100. The young Gauss produced the correct answer within seconds by a flash of mathematical insight, to the astonishment of his teacher and his assistant Martin Bartels. Gauss had realized that pairwise addition of terms from opposite ends of the list yielded identical intermediate sums: 1 + 100 = 101, 2 + 99 = 101, 3 + 98 = 101, and so on, for a total sum of 50 × 101 = 5050 ( ref. )

Fortunately 100 is an even number, so half of it -- 50 -- is how many pair-sums of 101 there are. The same formula works for odd numbers as well, but when you take half of an odd number, you get a remainder of ½.

For example, sum the numbers from 1 to 11: When you start pairing up the numbers, you get sums 1+11, 2+10, etc -- all of which equal 12 -- but these pairs are taken a total of "5½ times".

The nice thing about the formula n(n+1)/2 is that either n or n+1 must be even, so one or the other is always divisible by 2 -- you never actually have to deal with fractions.

In Kendor's original problem, the sum of 1 through 25 is 25*26/2 = 25*13.