My Math Forum Sequence problem

 Algebra Pre-Algebra and Basic Algebra Math Forum

April 16th, 2015, 11:44 AM   #1
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Sequence problem

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April 16th, 2015, 03:14 PM   #2
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Quote:
 $\displaystyle A \ \ number \ \ sequence \$ $\displaystyle \ \{a_n\}, \ \ \ (n = 1, \ 2, \ 3, \ ...) \ \ satisfies \ \ the \ \ following \ \ conditions.$ Express $\displaystyle \ \{a_n\} \$ as a function of $\displaystyle \ n.$ $\displaystyle 3a_{n + 1} \ = \ 2a_n \ + \ 1, \ \ \ (n = 1, \ 2, \ 3, \ ...), \ \ \ \ a_1 = 2.$
$\displaystyle a_1 = 2$

$\displaystyle 3a_2 = 2(2) + 1 = 5$

$\displaystyle 3a_2 = 5$

$\displaystyle a_2 = \dfrac{5}{3}$

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
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Hello, matisolla!

Quote:
 5) A number sequence $\{a_n\}$ satisfies the following conditions: $\;\;\;\;3a_{n+1} \,=\,2a_n\,+\,1,\;\;a_1=2$ Express $a_n$ as a function of $n.$

$\text{We have: }\:a_1 \,=\, 2,\; a_2 \,=\, \frac{5}{3},\; a_3 \,=\, \frac{13}{9}$

$\text{W\!e are given: }\;\;\;3a_{n+1} \;=\; 2a_n\,+\,1 \;\;\;\;(1)$
$\text{Next equation: }\; 3a_{n+2} \;=\; 2a_{n+1}\,+\,1\;(2)$

$\text{Subtract (2)-(1): }\:3a_{n+2}\,-\,3a_{n+1} \;=\;2a_{n+1}\,-\,2a_n$

$\;\;\;\;3a_{n+2}\,-\,5a_{n+1}\,+\,2a_n \;=\;0$

$\text{Let }x^n\,=\,a_n.$

$\text{W\!e have: }\:3x^{n+2}\,-\,5x^{n+1} \,+\,2x^n \;=\;0$

$\text{Divide by }x^n:\;\;3x^2\,-\,5x\,+\,2 \;=\;0$

$\;\;\;\;(x\,-\,1)(3x\,-\,2) \:=\:0 \;\;\;\Rightarrow\;\;\;x \:=\:1.\:\frac{2}{3}$

$\text{Assume }f(n)\text{ is of the form: }\:A(1^n)\,+\,B\left(\frac{2}{3}\right)^n$

$\;\;\;\;\begin{array}{cccccccc}f(1) = 2: & A \,+\,\frac{2}{3}B &=& 2 \\
f(2) = \frac{5}{3}: & A\,+\,\frac{4}{9}B &=& \frac{5}{3} \end{array}$

$\text{Solve the system: }\:A \,=\,1,\;B \,=\,\frac{3}{2}$

$\text{Therefore: }\:f(n) \;=\;1(1^n)\,+\,\frac{3}{2}\left(\frac{2}{3}\right )^n \;\;\;\Rightarrow\;\;\; f(n) \;=\;1\,+\,\left(\frac{2}{3}\right)^{n-1}$

May 24th, 2015, 05:17 AM   #4
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Quote:
 Originally Posted by soroban Hello, matisolla! $\text{We have: }\:a_1 \,=\, 2,\; a_2 \,=\, \frac{5}{3},\; a_3 \,=\, \frac{13}{9}$ $\text{W\!e are given: }\;\;\;3a_{n+1} \;=\; 2a_n\,+\,1 \;\;\;\;(1)$ $\text{Next equation: }\; 3a_{n+2} \;=\; 2a_{n+1}\,+\,1\;(2)$ $\text{Subtract (2)-(1): }\:3a_{n+2}\,-\,3a_{n+1} \;=\;2a_{n+1}\,-\,2a_n$ $\;\;\;\;3a_{n+2}\,-\,5a_{n+1}\,+\,2a_n \;=\;0$ $\text{Let }x^n\,=\,a_n.$ $\text{W\!e have: }\:3x^{n+2}\,-\,5x^{n+1} \,+\,2x^n \;=\;0$ $\text{Divide by }x^n:\;\;3x^2\,-\,5x\,+\,2 \;=\;0$ $\;\;\;\;(x\,-\,1)(3x\,-\,2) \:=\:0 \;\;\;\Rightarrow\;\;\;x \:=\:1.\:\frac{2}{3}$ $\text{Assume }f(n)\text{ is of the form: }\:A(1^n)\,+\,B\left(\frac{2}{3}\right)^n$ $\;\;\;\;\begin{array}{cccccccc}f(1) = 2: & A \,+\,\frac{2}{3}B &=& 2 \\ f(2) = \frac{5}{3}: & A\,+\,\frac{4}{9}B &=& \frac{5}{3} \end{array}$ $\text{Solve the system: }\:A \,=\,1,\;B \,=\,\frac{3}{2}$ $\text{Therefore: }\:f(n) \;=\;1(1^n)\,+\,\frac{3}{2}\left(\frac{2}{3}\right )^n \;\;\;\Rightarrow\;\;\; f(n) \;=\;1\,+\,\left(\frac{2}{3}\right)^{n-1}$

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: 7,674 Thanks: 2654 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. Thanks from matisolla

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### 2,5/3,13/9,35/27,97/81 sequence

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