Find $n$th iterative term in recurrence relation $a_{n+2}-5a_{n+1}+6a_{n}=0$

Question

The sequence $(a_{n})_{n\in \mathbb{N}}$ is given by recurrence relation : $$a_{1}=0,a_{2}=-6,$$ $$a_{n+2}-5a_{n+1}+6a_{n}=0\ \ (n\ge 1).$$

We get $$a_{3}=-30$$ $$a_{4}=-114$$ $$...$$

How to find $n$th iterative term $a_{n}$?

Answer 1

Since we have $$a_{n+2}-3a_{n+1}=2(a_{n+1}-3a_{n})$$ and $$a_{n+2}-2a_{n+1}=3(a_{n+1}-2a_{n}),$$ we have $$a_{n+1}-3a_n=2(a_{n}-3a_{n-1})=\cdots=2^{n-1}(a_2-3a_1)=-6\cdot 2^{n-1}$$and$$a_{n+1}-2a_{n}=3(a_n-2a_{n-1})=\cdots=3^{n-1}(a_2-2a_1)=-6\cdot 3^{n-1}.$$ Subtracting the latter from the former gives you $$-a_n=-6\cdot 2^{n-1}+6\cdot 3^{n-1},$$ i.e. $$a_n=6\cdot 2^{n-1}-6\cdot 3^{n-1}\ \ \ (n\ge 1).$$

Answer 2

Recurrence relations with constant coefficients can be solved in much the same way linear odes with constant coefficients can be solved. For a second "order" one as you have, suppose the solution is given by $$a_n = c_1 \lambda_1^n + c_2 \lambda_2^n$$ If you plug a possible solution of the form $c \lambda^n$ in, you will find an equation for the unknowns. You can see many techniques at the wiki for Recurrence Relations.

Answer 3

This is technically a differential equation:

$f(x+2)-5f(x+1)+6f(x)=0$

With simple differential equations we obtain,

$f(x)=c_{1}2^{x}+c_{2}3^{x}$

Where $c_{1}$ and $c_{2}$ are some constants. We don't know what these are but you given me initial conditions to solve the problem with.

$f(3)=-30$

$f(4)=-114$

We have,

$8c_{1}+27c_{2}=-30$

$16c_{1}+81c_{2}=-114$

With simple algebra we obtain, $c_{1}=3$ and $c_{2}=-2$.

So we obtain,

$a_{n}=3(2^{n})-2(3^{n})$

Hope this helped.

Sincerely, Aleksandar