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Yes, thank you.
Good proof! Thanks a lot
is very useful for calculating sums.
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
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.
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:
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!
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