Solve the recurrence $a_n=7a_{n-1}-10a_{n-2}$ where $a_{0}= 3$ and $a_{1}=3$

Question

Solve the recurrence $a_n=7a_{n-1}-10a_{n-2}$ where $a_{0}= 3$ and $a_{1}=3$

how can a $a_{0}$ and $a_{1}$ both equal $3$?

Answer 1

The recurrence states something that is implicit and may be the cause of your confusion:

The recurrence relation,

$$a_n = 7a_{n-1} - 10 a_{n-2}$$

Holds for all values of $n$ greater than $1$.

The equation says nothing about $a_0$ and $a_1$, so $a_0$ and $a_1$ can be anything. In your case, they equal $3$, and you can calculate $a_2$ by using the equation:

$$a_2 = 7a_1-10a_0=21-30=-9.$$

Answer 2

Strategy for this type of relation:

Solve $x^2 = 7x -10$

You get $x=2$ and $x=5$.

Then your solution must be in the form $a_n = c(2)^n+d(5)^n$, and you get $c$ and $d$ by substituting the initial values. So $3 = c+d$ and $3 = 2c+5d$. Solve this linear system.

I would explain why this works, but that would involve linear algebra.

The initial values are defined independently of the recurrence relation, since you can't use the relation to guess what $a_1$ and $a_2$ are.