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 March 9th, 2009, 11:18 PM #1 Senior Member   Joined: Sep 2008 Posts: 199 Thanks: 0 Binomial theorem By assuming$0<\sqrt{5}-2<\frac{1}{4}$ , deduce that the difference between $(\sqrt{5}+2)^5$ and an integer is less than $\frac{1}{1024}$ I need some hints to work on this . Thanks
 March 10th, 2009, 04:26 AM #2 Global Moderator   Joined: Dec 2006 Posts: 20,484 Thanks: 2041 Hint: show that $(\sqrt{5}+2)^5\,-\,(\sqrt{5}-2)^5\,=\,2(5(\sqrt5)^4(2)\,+\,10(\sqrt5)^2(2)^3\,+ \,2^5)\,=\,1364.$
March 10th, 2009, 06:29 AM   #3
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Re:

Quote:
 Originally Posted by skipjack Hint: show that $(\sqrt{5}+2)^5\,-\,(\sqrt{5}-2)^5\,=\,2(5(\sqrt5)^4(2)\,+\,10(\sqrt5)^2(2)^3\,+ \,2^5)\,=\,1364.$

Thanks skipjack , is it enough just by showing that .

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