the answer seems too odd for me -n(n+1)(2n+1)/6, and what mathematical induction can do is just to prove other than to derive. Hoping some genius could handle this out, with ease and simplisity, like the derivation of 1+a+a²+a³+...+an
it's a huge challenge!! Cuz Google doesn't provide a derivation!!:P
I don't see why you don't want it to be using induction, but whatever. I remember that question being asked here before, so there'll be a topic here somewhere with a proof. I remember that it has an induction proof and a different proof as well, so you get two answers!
I'll try to find the link now. Hang on.
Edit: That was easier than I thought! Here it is.
Last edited by mathsyperson (2006-03-13 01:43:42)
Why did the vector cross the road?
It wanted to be normal.
Thank you for the link. The link finally reminded me of the proof i once read in Thomas' Calculus book, it uses a neighbor elimation trick:
key thought -- (k+1)³-k³ =3k²+3k+1
3(1 +2 +3 +... +n)
+) 1 +1 +1 +... +1
3x + 3n(n+1)/2+n=n³+3n²+3n
3x=n³+3/2 n²+1/2 n= 1/2 n(2n²+3n+1)
x= 1/6 n (2n+1) (n+1)
There's also and binomial proof, which is more usable and universal, but it's harder too.
IPBLE: Increasing Performance By Lowering Expectations.
Good proof! Thanks a lot
a small error, though
f(x) = x^i should be x^k instead, same to the following
in order to know sum of i^k, we need to do know k-1 sums of i^l ,for l=1,2,...k-1
but to me, sum of squares is enough to derive Spearman's Coefficient
Last edited by George,Y (2006-03-19 14:51:28)