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

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

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

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

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

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

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

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

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