Whom of you can explain it to me? http://jeremykun.files.wordpress.com/2011/06/triangle-proof.png
If it's fairly easy and you think I am not pushing myself please do tell me, instead of giving out the answer.
BTW: Nice site.
@Ishaan94 - do you know how to get the sum of all the components in the last triangle?
No ._.
wait
Is it like this? (2n+1)+2(2n+1)+3(2n+1)+...+n(2n+1)
exactly - now factor out the (2n+1)
i think like this way: i will assume that this is true until line n+1 (go downward from first line), i have: first line is true, 1+n+n = 2n + 1. so with line: n+1 we have: (n+1) + ... (n+1) at first triangle, at second triangle, we have: 1+ 2 + ... + n+ n+1, and at third: n+1 + n + ... + 1, sum last line, we still get: 2(n+1) + 1 + .... 2(n+1) + 1
\[\frac{n\left(n+1\right) \left(2n+1\right)}2\]What about the six?
that is correct - now you have to subtract the sum of the previous 2 triangles from this
but why?
what you are trying to find is the sum of the numbers in the 1st triangle
\[1^1+2^2+3^3+...\]
that should have started with \(1^2\)
it is a visual proof of this sum
oh I see it
It's an interesting proof - I suggest you try and do it yourself - you'll have greater pleasure from that :)
thanks, i will try my best.
let me know if you require a clue.
when you do "see" how to do it - it is quite remarkable!
@Ishaan94 - try turning your head as you view the triangles on the left...
this really is a "visual" solution - so don't think about any complex math here
OMG this is the best proof I have seen in my life so far
:) glad you "saw" it at last!
it is a very pleasing proof
it is. and thanks a lot asnaseer. :-)
yw :)
really beautiful xD
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