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 Pre-Calculus Pre-Calculus Math Forum

 November 16th, 2014, 05:44 PM #1 Newbie   Joined: Sep 2014 From: Miami Posts: 18 Thanks: 2 Series Exponent Hello, I am starting with series and I have an example from the book but I cannot figure out what they did with the exponents. Can anyone help? I want to know how to go from the first expression to the second one. $\displaystyle \sum_{n=1}^{\infty}\frac{4^{n}}{3^{n-1}}=\sum_{n=1}^{\infty}4(\frac{4}{3})^{n-1}$ Thanks November 16th, 2014, 06:06 PM #2 Newbie   Joined: Sep 2014 From: Miami Posts: 18 Thanks: 2 Hello Admins, I am so sorry. I posted this here. I was looking for the LaTex and I posted without noticing I was in the incorrect section. Can you move it? Thanks November 16th, 2014, 06:24 PM #3 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra $$4^n = 4 \times 4^{n-1}$$so$$\frac{4^n}{3^{n-1}} = 4\times \frac{4^{n-1}}{3^{n-1}} = 4 \left( \frac43 \right)^{n-1}$$ November 16th, 2014, 06:32 PM   #4
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Joined: Sep 2014
From: Miami

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Quote:
 Originally Posted by v8archie $$4^n = 4 \times 4^{n-1}$$so$$\frac{4^n}{3^{n-1}} = 4\times \frac{4^{n-1}}{3^{n-1}} = 4 \left( \frac43 \right)^{n-1}$$
Thanks a lot. So simple lol This series and sequences are killing me  January 9th, 2015, 08:36 PM #5 Newbie   Joined: Jan 2015 From: usa Posts: 1 Thanks: 0 Have you tried graphing it yet? Tags exponent, series Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post bml1105 Algebra 12 July 27th, 2014 11:40 AM Kiphyl Algebra 5 December 30th, 2013 11:09 PM Thinkhigh Calculus 3 March 2nd, 2012 06:42 AM e81 Algebra 3 May 20th, 2011 01:46 PM greg1313 Applied Math 3 November 17th, 2009 07:39 PM

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