Question

Sum of first n triangle numbers

Answer 1

How do we find this?

Answer 2

1,3,6,10.. is this the series?

Answer 3

yeah, the one generated by n(n+1)/2

Answer 4

right - i remember that now
but i don't recall the sum formula

Answer 5

i can only suggest googling it

Answer 6

http://mathforum.org/library/drmath/view/56926.html

Answer 7

@UnkleRhaukus - good drawing

Answer 8

Yeah, but that's only proving what we get, not how to get there

Answer 9

yes - i see what you mean - it seems that its a guess, which is then proved by induction

Answer 10

\[S_n=\Sigma \frac{n(n+1)}{2} = \frac{1}{2} \Sigma (n^2+n) = \frac{1}{2}(\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2})\]
\[ =\frac{1}{12}(n(n+1)(2n+1)+3n(n+1)) = \frac{1}{12} (n(n+1)(2n+1+3)) \]
\[= \frac{1}{12}(n(n+1)(2n+4)) =\frac{2}{12}(n(n+1)(n+2)) = \frac{n(n+1)(n+2)}{6}\]

Answer 11

How did you get the sum of n^2?