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apple_pi Group TitleBest ResponseYou've already chosen the best response.0
How do we find this?
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
1,3,6,10.. is this the series?
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
yeah, the one generated by n(n+1)/2
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
right  i remember that now but i don't recall the sum formula
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
i can only suggest googling it
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.0
dw:1345110210994:dw
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
http://mathforum.org/library/drmath/view/56926.html
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus  good drawing
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
Yeah, but that's only proving what we get, not how to get there
 2 years ago

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
yes  i see what you mean  it seems that its a guess, which is then proved by induction
 2 years ago

nightwill Group TitleBest ResponseYou've already chosen the best response.1
\[S_n=\Sigma \frac{n(n+1)}{2} = \frac{1}{2} \Sigma (n^2+n) = \frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})\] \[ =\frac{1}{12}(n(n+1)(2n+1)+3n(n+1)) = \frac{1}{12} (n(n+1)(2n+1+3)) \] \[= \frac{1}{12}(n(n+1)(2n+4)) =\frac{2}{12}(n(n+1)(n+2)) = \frac{n(n+1)(n+2)}{6}\]
 2 years ago

apple_pi Group TitleBest ResponseYou've already chosen the best response.0
How did you get the sum of n^2?
 2 years ago
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