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This is a very interesting word problem that I came across in an old textbook of mine. So I know its got something to do with sequencing and series, which yields the shortest, simplest proofs, but other than that, the textbook gave no hints really and I'm really not sure about how to approach it. Any guidance hints or help would be truly greatly appreciated. Thanks in advance :) So anyway, here the problem goes:

Simplify $$\frac{1}{1\times2\times3} + \frac{1}{2\times3\times4} +\frac{1}{3\times4\times5}+\cdots+\frac{1}{n(n+1)(n+2)}$$

anonymous
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  • You use the same phrasing you did in your other question that constitutes an attemp to cheat at Australian Junior Math Contest (currently running). Is this an attempt to cheat as well? – Jyrki Lahtonen May 30 '17 at 06:41

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You just need to realize that $$\frac{1}{1\times2\times3}=\frac12(\frac{1}{1\times2}-\frac{1}{2\times3})\\ \frac{1}{2\times3\times4}=\frac12(\frac{1}{2\times3}-\frac{1}{3\times4})\\$$ Now combine everything, you can have most of the terms cancelled. $$\frac{1}{1\times2\times3} + \frac{1}{2\times3\times4} +\frac{1}{3\times4\times5}+\cdots+\frac{1}{n(n+1)(n+2)}=\frac12(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\cdots-\frac{1}{(n+1)(n+2)})\\=\frac12(\frac{1}{1\times2}-\frac{1}{(n+1)(n+2)})\\=\frac12(\frac{(n+1)(n+2)-2}{2(n+1)(n+2)})\\=\frac{n(n+3)}{4(n+1)(n+2)}$$

Julia
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