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 March 16th, 2015, 12:46 PM #1 Newbie   Joined: Mar 2015 From: chicago Posts: 2 Thanks: 0 General term of a sequence Hi, I'm looking for the general term of this sequence $a_0= 3$ $a_{n+1} = 2 a_n + 5^n$ I started to find an obvious solution but i couldn't. Any suggestion would be appreciated Last edited by Aaron; March 16th, 2015 at 12:50 PM.
 March 16th, 2015, 01:04 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Look for a linear recurrence relation. It should be of order 2: a(n) = x*a(n-1) + y*x(n-2). In this case there's also a closed form a(n) = $\alpha\cdot\beta^n+\gamma\cdot\delta^n.$
March 16th, 2015, 03:08 PM   #3
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Hello, Aaron!

Quote:
 I'm looking for the general term of this sequence: $\;\;\;a_0\:=\: 3,\;\;a_{n+1} \:=\: 2 a_n\,+\,5^n$
$\begin{array}{ccccc} (1) & a_{n+1} &\;=\;& 2a_n\,+\,5^n
\\ (2) & a_{n+2} &\;=\;& 2a_{n+1}\,+\,5^{n+1} \end{array}$

$\text{Subtract [2] - [1]: }$
$\;\;\;a_{n+2}\,-\,a_{n+1} \;=\;2a_{n+1}\,-\,2a_n \,+\,5^{n+1}\,-\, 5^n$

$\;\;a_{n+2}\,-\,3a_{n+1}\,+\,2a_n \;=\;4\cdot 5^n$

$\text{The characteristic equation is: }\:X^2\,-\,3X\,+\,2 \:=\:0 \;\;\;\Rightarrow\;\;\;X \:=\:1,\,2$

$\text{I conjecture that the function is: }\:f(n) \:=\:A\,+\,B\cdot2^n\,+\,C\cdot5^n$

$\text{The first few terms are: }\:3,\,11,\,47,\,219,\,1063$

$\begin{array}{cccccccc} f(0) = 3: & A\,+\,B\,+\,C &=& 3 \\
f(1) = 11: & A\,+\,2B\,+\,5C &=& 11 \\
f(2) = 47: & A\,+\,4B\,+\,26C &=& 47 \end{array}$

$\text{Solve the system: }\:A= 0,\;B = \frac{4}{3},\;C = \frac{5}{3}$

$\text{The function is: }\:f(n) \;=\;\frac{4}{3}\,2^n\,+\,\frac{5}{3}\,5^n \;\;\;\Rightarrow\;\;\; f(n) \;=\;\frac{1}{3}\left(2^{n+2}\,+\,5^{n+1}\right)$

 March 16th, 2015, 03:15 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs You could also use the inhomogeneous difference equation: $\displaystyle a_{n+1}-a_{n}=5^n$ You then see that the homogeneous solution is: $\displaystyle h_n=c_1\cdot2^n$ And the particular solution must take the form: $\displaystyle p_n=A\cdot5^n$ Use the method of undetermined coefficients to find $A$, then use the initial value to find the parameter $c_1$. You should find: $\displaystyle a_n=\frac{2^{n+3}+5^n}{3}$ Thanks from Country Boy
 March 16th, 2015, 08:00 PM #5 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms soroban and Mark are correct. You should also check that the formula fits the recurrence, just in case.
 March 17th, 2015, 03:59 PM #6 Newbie   Joined: Mar 2015 From: chicago Posts: 2 Thanks: 0 Thank for the help, I found a different result $a_{n+2}=2a_{n+1} + 5^{n+1}$ $a_{n+1} - 2a_n = 5^n$ $a_{n+2}=2a_{n+1} + 5(a_{n+1}-2a_n)$ $a_{n+2}=7a_{n+1}-10a_n$ Characteristic equation : $r^2 - 7r +10 =0$ $r_1=2$ and $r_2=5$ General term : $a_n= \alpha r_1^n + \beta r_2^n$ The first 2 terms give us the coefficients : $a_0 = 3$ and $a_1 = 7$ $\alpha = \frac{8}{3}$, $\beta = \frac{1}{3}$ $a_n = \frac{8}{3}.2^n +\frac{1}{3}.5^n$
March 17th, 2015, 05:06 PM   #7
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Quote:
 Originally Posted by Aaron Thank for the help, I found a different result... $a_n = \frac{8}{3}.2^n +\frac{1}{3}.5^n$
That's just a slightly different form for the same closed expression I gave.

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