# Find the solution to the recurrence relation $a_{n} = 3a_{n−1} +4a_{n−2}$ with initial terms $a_{0}=5$ and $a_{1} =8$

For the question

Find the solution to the recurrence relation $$a_{n} = 3a_{n−1} +4a_{n−2}$$ with initial terms $$a_{0}=5$$ and $$a_{1} =8$$

I think the way to solve this is the Characteristic Root Technique since the recurrence relation is a combination of two previous terms.

So the characteristic polynomial is $$x^2 + 3x + 4 =0$$. In the example solutions I have seen the two distinct characteristic roots should be easy to find, but for this question, I don't know how to move forward. If I do completing the square then I get $$(x+\frac{3}{2})^2 + \frac{7}{4}=0$$.

I am stuck here. How do I find the solution?

• Do you remember the quadratic formula from Algebra I? – amd Mar 16 '19 at 18:17

For the given recurrence relation $$a_{n} -3a_{n−1} -4a_{n−2}=0$$, the characteristic polynomial is $$x^2 - 3x - 4 =0$$, so we have to solve $$x^2 - 3x - 4=\left(x-\frac{3}{2}\right)^2 - \left(\frac{5}{2}\right)^2=(x+1)(x-4)=0.$$ Therefore $$a_n=C_1 \cdot 4^n+ C_2\cdot (-1)^n$$ where $$C_1$$ and $$C_2$$ are constants to be determined. Now find such constants by using the initial terms $$a_{0}=5$$ and $$a_{1} =8$$.