What is $a_5$, given the recurrence $a_{n+1}=a_n+2a_{n-1}$ and we know that $a_0 = 4, a_2 =13$ I am having a very hard time figuring this out.
So far I have been able to do the following:
Writing the recurrence as a characteristic polynomial = $x^2-x-2=0$ so there are roots, $x=2, x=-1$.
So there will be the general solution of $X_n = c_12^n + c_2 n(-1)^n$
I think the first constant will be 4..but I really am not sure.
And from there on I am very confused with where to go.  I have a big feeling this sort of question will be on my test in a few weeks so an explanation would be very helpful :)
Thanks in advance!
 A: We know that $a_{n+1}=a_n+2a_{n-1}.$
So, $13= a_1 +2*4$. This implies that $a_1=5$.
We now just use the recurrence relation.
$a_3 = 13 + 2*5= 23$
$a_4=23 +2*13=49$
$a_5= 49 + 2*23= 95$. That finishes it. Just a simple calculation!
A: Direct calculation is quite a bit easier.
However, let us look into the characteristic equation method, since the technique will undoubtedly be needed later.  The roots are $2$ and $-1$, so the general solution is $A(2^n)+B(-1)^n$.
We need to find $A$ and $B$.
Setting $n=0$, we find that $A(2^0)+B(-1)^0= 4$. So $A+B=4$.
Setting $n=2$, we get $A(2^2)+B(-1)^2=13$, so $4A+B=13$. 
Solve the system of two equations in two unknowns. We get $A=3$ and $B=1$. Thus 
$$a_n=3(2^n)+(-1)^n.$$
Setting $n=5$, we get $96-1$, that is, $95$.  
Remark: More directly, $a_2=a_1+2a_0=13$, and therefore $a_1=5$. Now we can use the recurrence to find $a_3$ then $a_4$ then 4a_5$.  
A: You know $a_0$ and $a_2$ so $13=a_1+2(4)$ and $a_1=5$. Then $a_3=13+2(5)=23$. Then $a_4=23+2(13)=49$. Then $a_5=49+2(23)=95$.
Alternatively, you can solve the recurrence relation explicitly via generating functions knowing $a_0$ and $a_1$.
