# The sequence $(a_0,a_1,a_2,\cdots,)$ satisfies $a_{n+1}=a_n+2a_{n−1}$. What is $a_5$?

Assume that the sequence $(a_0,a_1,a_2,\cdots,)$ satisfies the recurrence $\displaystyle a_{n+1}=a_n+2a_{n−1}$. We know that $a_0=4$ and $a_2=13$. What is $a_5$?

I got $a_1=5, a_3=23, a_4=49, a_5=95$

• Is this formatted correctly? Is it supposed to be $a_{n+1}=a_n+2a_{n-1}$ instead? – steve Sep 1 '14 at 21:38
• The values you computed look correct. What exactly is your question? – JimmyK4542 Sep 1 '14 at 21:38
• Hint. – user153012 Sep 2 '14 at 0:02

If you want to calculate the $n$-th term, here is how:

$a_{n+1} - a_n - 2a_{n-1} = 0 \to x^2 - x - 2 = 0 \to (x-2)(x+1) = 0 \to x = -1, 2$. Thus the

$a_n = A(-1)^n + B2^n$. We have that: $a_0 = 4 \to a_1 = a_2 - 2a_0 = 13 - 2(4) = 5$. So:

$5 = a_1 = A(-1)^1 + B2^1 = -A + 2B$, and

$13 = a_2 = A(-1)^2 + B2^2 = A + 4B$.

Adding these equations we find $B = 3$, and solve for $A = 13 - 4(3) = 1$. Thus:

$a_n = (-1)^n + 3\cdot 2^n$.

Check that $a_5 = (-1)^5 + 3\cdot 2^5 = -1 + 96 = 95$ !