My Math Forum Series Exponent

 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: 7,638 Thanks: 2623 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
Newbie

Joined: Sep 2014
From: Miami

Posts: 18
Thanks: 2

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?

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