My Math Forum geometric series; find nth term

 Algebra Pre-Algebra and Basic Algebra Math Forum

 February 26th, 2014, 11:35 AM #1 Member   Joined: Sep 2013 Posts: 75 Thanks: 0 geometric series; find nth term 8x^2 + 4x^3 + 2x^4 Find nth term? T2÷T1 = T3÷T2 = r I got 1/2x = r; then I stuck!
February 26th, 2014, 12:32 PM   #2
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 408

Re: geometric series; find nth term

Hello, bongantedd!

Are you familiar with Geometric Series formulas?

Quote:
 $8x^2\,+\,4x^3\,+\,2x^4\,+\,\cdots \;\;\text{Find }n^{th}\text{ term}$ $T_2\,\div\,T_1 \:=\: T_3\,\div\,T_2 \:=\: r \;\;\text{I got: }\,r\,=\,\frac{1}{2}x$ [color=beige]. . . [/color][color=blue]Correct! ~~ Keep going.[/color]

$\text{The }n^{th}\text{ term is given by: }\:a_n \:=\:a_1\cdot\,r^{n-1}$

$\text{W\!e have: }\:a_1 \,=\,8x^2,\;r \,=\,\frac{x}{2}$

$\text{Hence: }\:a_n \;=\;8x^2\left(\frac{x}{2}\right)^{n-1} \;=\;8x^2\left(\frac{x^{n-1}}{2^{n-1}}\right) \;=\;\frac{2^3}{2^{n-1}}\left(x^2\,\cdot\,x^{n-1}\right)$

$\text{Therefore: }\:a_n \;=\;\frac{x^{n+1}}{2^{n-4}}$

 February 26th, 2014, 04:38 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 If the series were just x^2 + x^3 + x^4, etc., one could just write the answer almost instantly.

 Tags find, geometric, nth, series, term

### geometric sequence of 8x 4x 2x

Click on a term to search for related topics.
 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Polaris84 Real Analysis 3 October 26th, 2013 02:12 PM wilhelm Applied Math 1 November 28th, 2012 06:00 AM The Chaz Real Analysis 11 February 7th, 2011 04:52 AM ammarkhan123 Real Analysis 0 April 16th, 2010 07:52 AM Sean Algebra 4 January 29th, 2007 02:13 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top