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