Question

I am a little baffled by this formula
1+....+n = [n(n+1)]/2
I tried to figure it out, but I am left nowhere comprehensible. Would someone be so kind to explain how the formula came about without induction?

Answer 1

Um, wow.... I haven't the fuzziest. If I may ask, what course is this for?

Answer 2

"Intro to Mathematical analysis" in spivak's calculus text @austinL

Answer 3

honest to truth openstudy reintroduced me to my old interest in math

Answer 4

Pair the elements up in the following way.
\[1+2+...+n=(1+n)+(2+(n-1))+...x\]Where \(x\) is either \(\frac{n+1}{2}\) if \(n\) is odd, and \((\frac{n}{2}-1)+(\frac{n}{2}+1)=n\) if \(n\) is even.

Each term (except \(x\)) is equal to \(n+1\), and there are \(n/2\) terms if \(n\) is even, immediately giving us the formula\[\frac{n(n+1)}{2}.\]If \(n\) is odd, then we have \(\frac{n-1}{2}\) terms, giving us\[\frac{(n+1)(n-1)}{2}+\frac{n+1}{2}=\frac{n^2+n}{2}=\frac{n(n+1)}{2}\]

Answer 5

I hope that was clear enough :/

Answer 6

Thanks, KG. I'll review what you typed later when I get home. The Safari browser in my iPhone is rendering the latex with errors

Answer 7

Well the premise can be done without latex. Basically divide the sum into pairs in the following way

1+2+...+n=(1+n)+(2+(n-1))+(3+(n-2))+...

Then you just have to consider the cases when n is even or odd.