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 June 3rd, 2012, 09:26 AM #1 Newbie   Joined: Jun 2012 Posts: 16 Thanks: 0 Basic Algebra, fraction of a fraction... Could someone explain to me how to get from $\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^n}}$ to $\frac{1}{3}(1+\frac{1}{n})$
 June 3rd, 2012, 11:58 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,319 Thanks: 936 Re: Basic Algebra, fraction of a fraction... You can't...unless you make an illegal u-turn!
 June 3rd, 2012, 12:03 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Basic Algebra, fraction of a fraction... $\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^n}}=\frac{n+ 1}{3^{n+1}}\cdot\frac{3^n}{n}=\frac{3^n}{3\cdot3^n }\cdot\frac{n+1}{n}=\frac{1}{3}$$\frac{n+1}{n}$$=\ frac{1}{3}$$1+\frac{1}{n}$$$
June 3rd, 2012, 03:13 PM   #4
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Re: Basic Algebra, fraction of a fraction...

Quote:
 Originally Posted by MarkFL $\frac{\frac{n+1}{3^{n+1}}}{\frac{n}{3^n}}=\frac{n+ 1}{3^{n+1}}\cdot\frac{3^n}{n}=\frac{3^n}{3\cdot3^n }\cdot\frac{n+1}{n}=\frac{1}{3}$$\frac{n+1}{n}$$=\ frac{1}{3}$$1+\frac{1}{n}$$$
Mark to the Rescue again!
Thanks man, big help!

 June 3rd, 2012, 03:51 PM #5 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,319 Thanks: 936 Re: Basic Algebra, fraction of a fraction... But but...Mark cheated! He used 3^(n+1) That looked to me like 3^n + 1
 June 3rd, 2012, 06:14 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 Math Focus: Calculus/ODEs Re: Basic Algebra, fraction of a fraction... I thought that was most likely the way you interpreted that.

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