Here's the question you clicked on:
apple_pi
Sum of first n triangle numbers
1,3,6,10.. is this the series?
yeah, the one generated by n(n+1)/2
right - i remember that now but i don't recall the sum formula
i can only suggest googling it
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@UnkleRhaukus - good drawing
Yeah, but that's only proving what we get, not how to get there
yes - i see what you mean - it seems that its a guess, which is then proved by induction
\[S_n=\Sigma \frac{n(n+1)}{2} = \frac{1}{2} \Sigma (n^2+n) = \frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})\] \[ =\frac{1}{12}(n(n+1)(2n+1)+3n(n+1)) = \frac{1}{12} (n(n+1)(2n+1+3)) \] \[= \frac{1}{12}(n(n+1)(2n+4)) =\frac{2}{12}(n(n+1)(n+2)) = \frac{n(n+1)(n+2)}{6}\]
How did you get the sum of n^2?