discrete mathematics - Solve the following recurrence relation: $a_{n}=10a_{n-2}$

I'm trying to solve the following recurrence equation $a_{n}=10a_{n-2}$.

Initial conditions: $a_0=1$, $a_1=10$

I have tried to use characteristic polynomial and generating functions but both methods lead to contradiction.

I think I should observe $\mathbb{N}_{even}$ and $\mathbb{N}_{odd}$ as different cases but I don't know how to do it formally.

Any help would be appreciated.

Answer

We need initial values for $a_0$ and for $a_1$ to start with.

Notice that, $$a_2 = 10a_0$$ , $$a_4 = 10a_2 =10^2a_0$$ $$a_6=10a_4 =10^3 a_0$$

With the given initial values we get $$a_{2k}=10^ka_0 = 10^k$$

Similarly you get $$a_{2k+1}=10^{k+1}$$