Another "Can u add up?????" thread

What is the sum of all integers from 1 to 100 (or any integer to any other integer)?

There is a much quicker way than going 1+2+3... etc.

This would involve deriving a general equation so you can work it out for 1,000 or 1,000,000 without any trouble at all.

This is approximately grade 10 math.
The simple solution is..
(1+ 2+ 3 +4 +5+...+100)+
(100+99 +98+.... +1)=
101+101+.... +101= 100*101=10100
But we counted each number twice so divide by 2.
10100/2=5050

Similiarly 1+2+...+n=n*(n+1)/2

(m+1)+(m+2)+...+n= n*(n+1)/2 - m*(m+1)/2

The story is that Guass (Arguably the greatest mathematician ever) was asked this by his teacher when he was about 7 (with a goal of keeping him and the rest of his class quite for the day). He wrote down the answer to the teachers annoyance and amazement.

Originally posted by Virak
This is approximately grade 10 math.
The simple solution is..
(1+ 2+ 3 +4 +5+...+100)+
(100+99 +98+.... +1)=
101+101+.... +101= 100*101=10100
But we counted each number twice so divide by 2.
10100/2=5050

Similiarly 1+2+...+n=n*(n+1)/2

(m+1)+(m+2)+...+n= n*(n+1)/2 - m*(m+1)/2

The story is that Guass (Arguably the ...[text shortened]... his class quite for the day). He wrote down the answer to the teachers annoyance and amazement.

Correct apart from the slight mistake in the second part. You added from (m+1) to n, should be m to n:

m+(m+1)+(m+2)+...+n = (n*(n+1)/2) - (m*(m+1)/2) + m

Which could be simplified to:

http://www.lunacy.force9.co.uk/images/gauss.jpg

The easiest way to remember the formula is:
Average number times number of entries i.e. (m+n)/2 * (m - n + 1)

Added from M+1 on purpose.
The idea being to add up to n and subtract the numbers from 1 to m