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 March 14th, 2011, 06:13 AM #1 Senior Member   Joined: Dec 2010 Posts: 233 Thanks: 0 help binomial expansion 1.without using tables or calculator,evaluate ( ?7+?3)^6+(?7-?3)^6. 2.Obtain the binomial expansion of ( ?2+1)^5 in the form m?2 + n, where m and n are integers.State the corresponding result for the expansion of (?2-1)^5,and show that ( ?2-1)^5 is the reciprocal of ( ?2+1)^5. Please help me to solve these question.I need help.i am weak at math.
 March 14th, 2011, 09:22 AM #2 Senior Member     Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 Re: help binomial expansion 1. Use this formula: a^3+b^3=(a+b)(a^2-ab+b^2) , for your case: (( ?7+?3)^2)^3+((?7-?3)^2)^3 , the result is 7040. 2. Just use: (a+b)^5=a^5+5a^4*b+10a^3*b^2+10a^2*b^3+5a*b^4+b^5 , so: (?2+1)^5 is 29?2+41 and: (?2-1)^5 is 29?2-41
 March 14th, 2011, 09:30 AM #3 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Above problems only requires basic algebra, and above average mathematical talent.
March 14th, 2011, 09:32 AM   #4
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Re: help binomial expansion

Quote:
 Originally Posted by hoyy1kolko . . . show that ( ?2-1)^5 is the reciprocal of ( ?2+1)^5.
$(29\sqrt{2}\,-\,41)\,\cdot\,\frac{29\sqrt{2}\,+\,41}{29\sqrt{2}\ ,+\,41}\,=\,\frac{1}{29\sqrt{2}\,+\,41}$

 March 14th, 2011, 09:41 AM #5 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,767 Thanks: 5 Re: help binomial expansion I hope I have what it takes...! $(\sqrt{7}+\sqrt{3})^6 + (\sqrt{7}-\sqrt{3})^6$ Using the above suggestion, we have $((\sqrt{7}+\sqrt{3})^2)^3 + ((\sqrt{7}-\sqrt{3})^2)^3$ I'll try the squaring first... $(10 + 2\sqrt{21})^3 + (10 - 2\sqrt{21})^3$ Then "a + b" is just 20. $a^2= 184 + 40\sqrt{21}$ $b^2= 184 - 40\sqrt{21}$ $ab= 100 - 84 = 16.$ So... (20)(368 - 16) = 20*352 = 7040
 March 16th, 2011, 09:30 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: help binomial expansion 2.) If we define: $P_n=2P_{n-1}+P_{n-2}$, $P_0=0,P_1=1$ $Q_n=2Q{n-1}+Q_{n-2}$, $Q_0=1,Q_1=1$ Then it can be shown that: $$$sqrt{2}\pm1$$^n=P_n\sqrt{2}\pm Q_n$ $P_2=2(1)+0=2$, $Q_2=2(1)+1=3$ $P_3=2(2)+1=5$, $Q_3=2(3)+1=7$ $P_4=2(5)+2=12$, $Q_4=2(7)+3=17$ $P_5=2(12)+5=29$, $Q_5=2(17)+7=41$ Thus: $$$sqrt{2}\pm1$$^5=29\sqrt{2}\pm41$ $\frac{1}{$$\sqrt{2}\pm1$$^n}\cdot\frac{$$\sqrt{2}\ mp1$$^n}{$$\sqrt{2}\mp1$$^n}=\frac{$$\sqrt{2}\mp1\ )^n}{\(2-1$$^n}=$$\sqrt{2}\mp1$$^n$

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