Given the recurrence relation $a_n = 2a_{n-1} - a_{n-2}$ how can I find the initial terms if $a_9 = 30$?

For which $x$ are there initial terms which make $a_9 = x$?

I know that there is no solution to the recurrence relation if $a_0 = 1$ and $a_1 = 2$ using the Characteristic Root Technique:

$ x^2 -2x +1 = 0$ which results in $r_1 = 1$ and $r_2 = 1$.

$a_n = ar_1^n + br_2^n$

$a_n = a(1)^n + b(1)^n$

$a_0 = a(1)^0 + b(1)^0$ which results in $1 = a + b$.

$a_1 = a(1)^1 + b(1)^1$ which results in $2 = a + b$.

Obviously $1 != 2$ resulting in no solution.



and in the end



This linear equation in two unknowns has an infinity of solutions.


With repeated roots, you're supposed to use $a_n = ar^n+bnr^n$. So if $a_0=1$ and $a_1=2$, then you have

$a_0 = 1a +0b = 1$

$a_1 = 1a + 1b = 2$

So $a=1$, $b = 1$ and $a_n = 1+n$.

You can check this with

$1+n = a_n = 2a_{n-1}-a_{a-2} = 2(1+n-1) - (1+n-2) = 2n-n+1= 1+n$

If you're just given $a_9=30$, that's insufficient to find the initial terms, as you have two unknowns but only one equation.

Also, if you're using MathJax, typing != will result in a space between ! and =. Use \neq instead.

  • $\begingroup$ Ahh yes I did use the wrong formula. Thank you for your response! Very helpful as I am preparing for my final. $\endgroup$ – Josh Garza May 8 '18 at 18:47

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