anonymous
  • anonymous
Whom of you can explain it to me? http://jeremykun.files.wordpress.com/2011/06/triangle-proof.png
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
http://mathoverflow.net/questions/8846/proofs-without-words/69756#69756
anonymous
  • anonymous
If it's fairly easy and you think I am not pushing myself please do tell me, instead of giving out the answer.
asnaseer
  • asnaseer
BTW: Nice site.

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asnaseer
  • asnaseer
@Ishaan94 - do you know how to get the sum of all the components in the last triangle?
anonymous
  • anonymous
No ._.
anonymous
  • anonymous
wait
anonymous
  • anonymous
Is it like this? (2n+1)+2(2n+1)+3(2n+1)+...+n(2n+1)
asnaseer
  • asnaseer
exactly - now factor out the (2n+1)
anonymous
  • anonymous
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
anonymous
  • anonymous
\[\frac{n\left(n+1\right) \left(2n+1\right)}2\]What about the six?
asnaseer
  • asnaseer
that is correct - now you have to subtract the sum of the previous 2 triangles from this
anonymous
  • anonymous
but why?
asnaseer
  • asnaseer
what you are trying to find is the sum of the numbers in the 1st triangle
asnaseer
  • asnaseer
\[1^1+2^2+3^3+...\]
asnaseer
  • asnaseer
that should have started with \(1^2\)
asnaseer
  • asnaseer
it is a visual proof of this sum
anonymous
  • anonymous
oh I see it
asnaseer
  • asnaseer
It's an interesting proof - I suggest you try and do it yourself - you'll have greater pleasure from that :)
anonymous
  • anonymous
thanks, i will try my best.
asnaseer
  • asnaseer
let me know if you require a clue.
asnaseer
  • asnaseer
when you do "see" how to do it - it is quite remarkable!
asnaseer
  • asnaseer
@Ishaan94 - try turning your head as you view the triangles on the left...
asnaseer
  • asnaseer
this really is a "visual" solution - so don't think about any complex math here
anonymous
  • anonymous
OMG this is the best proof I have seen in my life so far
asnaseer
  • asnaseer
:) glad you "saw" it at last!
asnaseer
  • asnaseer
it is a very pleasing proof
anonymous
  • anonymous
it is. and thanks a lot asnaseer. :-)
asnaseer
  • asnaseer
yw :)
rational
  • rational
really beautiful xD

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