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April 16th, 2015, 11:44 AM  #1 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Sequence problem
Please help me with this problem Im having a hard time with sequences and series!

April 16th, 2015, 03:14 PM  #2  
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  Quote:
$\displaystyle 3a_2 = 2(2) + 1 = 5$ $\displaystyle 3a_2 = 5$ $\displaystyle a_2 = \dfrac{5}{3}$ After this, $\displaystyle \ \ \dfrac{5}{3}$ will take on the role that $\displaystyle \ a_1 = 2 \ $ had above. Continue this process. Including $\displaystyle \ a_1 = 2, \ $ the first five numbers are: 2, 5/3, 13/9, 35/27, 97/81 I found this sequence difficult to determine a pattern, but I wrote them as this eventually: $\displaystyle \dfrac{2}{1}, \ \dfrac{5}{3}, \ \dfrac{13}{9}, \ \dfrac{35}{27}, \ \dfrac{97}{81} \ = $ $\displaystyle \dfrac{ 1 + 1}{1}, \ \dfrac{3 + 2}{3}, \ \dfrac{9 + 4}{9}, \ \dfrac{27 + 8}{27}, \ \dfrac{81 + 16}{81} $ Can you come up with a general formula for $\displaystyle \ a_n \ $ in terms of $\displaystyle \ n \ $ by studying that?  
April 16th, 2015, 05:37 PM  #3  
Math Team Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408  Hello, matisolla! Quote:
 
May 24th, 2015, 05:17 AM  #4 
Senior Member Joined: Jan 2015 From: USA Posts: 107 Thanks: 2  Thanx Soroban! This method looks very useful but I cant understand the theory behind it. Does this method of solving sequences have a name? I would like to read further about it. By the way, are you Japanese? Because I speak Japanese and "Soroban" is a very fitting name for someone like you lol 
May 24th, 2015, 10:14 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,634 Thanks: 2620 Math Focus: Mainly analysis and algebra 
Generally, a difference equation has solutions $A\alpha^n$ where $\alpha$ is a root of the characteristic polynomial (quadratic in the case above). The general solution is a linear combination of each of the terms (one for each root). One can also derive a solution using the first order equation. Suppose that $a_n = f(n)$ is a solution of $3a_{n+1}2a_n= 1$ and that $a_n = g(n)$ is a solution of the homogeneous equation $3a_{n+1}2a_n= 0$, Then $a_n = f(n) + cg(n)$ is also a solution of the nonhomogeneous equation (where $c$ is a constant). Thus we may seek a constant solution to the nonhomogeneous equation and add to it some multiple of the solution to the homogeneous equation. Our initial condition will fix the multiple. Thus we seek a constant $a$ such that $3a  2a = 1$ which implies $a=1$, and the solution of the nonhomogeneous equation $a_{n+1}={2 \over 3}a_n$ which is $g(n) = \left( {2 \over 3} \right)^n$. Thus our general solution is $a_n = 1 + c\left( {2 \over 3} \right)^n$ where $c$ is a constant determined by the initial condition. 

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