[SOLVED] Need help on this proof

**gigglie** · May 5th 2008, 11:23 AM

Prove that a[n]=x*2^n+y*3^n is a solution to the equation
a[n] = 5*a[n-1] - 6*a[n-2] when n>=2 ,x and y are integers and numbers in [] s are subscripts and ^ denotes power .

**Isomorphism** · May 5th 2008, 11:36 AM

Originally Posted by gigglie

Prove that a[n]=x*2^n+y*3^n is a solution to the equation
a[n] = 5*a[n-1] - 6*a[n-2] when n>=2 ,x and y are integers and numbers in [] s are subscripts and ^ denotes power .

To prove $a_n = x2^n+y3^n$ is a solution, substitute this in $a_n = 5a_{n-1}- 6a_{n-2}$ and see if it satisfies.

$a_n = 5a_{n-1}- 6a_{n-2} \Rightarrow 5(x2^{n-1}+y3^{n-1}) - 6(x2^{n-2}+y3^{n-2})$

$= (10x - 6x)2^{n-2}+(15y - 6y)3^{n-2} = (4x)2^{n-2}+(9y)3^{n-2} = a_n$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~~~
The standard way to solve these homogenus recurrences is to substitute $a_n = r^n$ and see what value of r you get.Then the general solution is a linear combination of the obtained $r^n$

$a_n = 5a_{n-1}- 6a_{n-2}$
$r^n = 5r^{n-1}- 6r^{n-2}$
$r^2 = 5r- 6 \Rightarrow r = 2,3$

Thus $a_n = x2^n+y3^n$

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