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  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




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Topic review (newest first)

2006-04-01 06:45:05

Yes, thank you.

2006-03-20 13:45:22

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 tongue

2006-03-19 02:31:06

The function:

is very useful for calculating sums.
That's because:

If you want to calculate this:

, you may express the f(x) = x^i as:
where r(x) is the "remainder", which has less power than f(x).
Example for f(x)=x^1:
S_2 (x) = (x+1)^2-x^2=x^2+2x+1-x^2=2x+1
(S_2 (x))/2=x+1/2
f(x)=x^1=x=0.5 S_2(x) - 0.5 = 0.5(S_2(x)-1).

There's also and binomial proof, which is more usable and universal, but it's harder too.

2006-03-14 14:53:32

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)-k =3k+3k+1

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


3x + 3n(n+1)/2+n=n+3n+3n

3x=n+3/2 n+1/2 n= 1/2 n(2n+3n+1)

x= 1/6 n (2n+1) (n+1) cool

2006-03-14 00:48:48

Many thanks! roll smile

2006-03-14 00:42:34

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.

2006-03-14 00:32:48

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+a+a+...+an

it's a huge challenge!! Cuz Google doesn't provide a derivation!!tongue

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