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4U Revising Game (3 Viewers)

harism

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omg! now i notice the minus.
great.

btw nice sol azureus.
 

azureus88

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lol gurmies, its kinda hard to type out a solution for that question. If you write it out in columns, a lot of stuff cancels out and you get:

[maths]\sum_{r=3}^{2001}\frac{8}{r(r-2)(r+2)}\\=\sum_{r=3}^{2001}(-\frac{2}{r}+\frac{1}{r-2}+\frac{1}{r+2})\\=(\frac{-2}{3}+\frac{1}{1}-\frac{2}{4}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\\=\frac{11}{12}+(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003})\\> \frac{11}{12}[/maths]

btw are you sure its less than and not greater than?
 

harism

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ha! trust lolokay for a solution like that.
;)
 

harism

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Why so silent? has everyone died?

lol okay.

you were meant to put a question up.

heres one.



 
Last edited:

Drongoski

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Edit:

Above solution amended; missed out 'x^n' in numerator
 
Last edited:

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