Find a solution of $a_{n+2} = 5a_{n+1} - 6a_n$ that satisfies the initial conditions $a_0 =3$ and $a_1 =11$

Question

$$a_{n+2} = 5a_{n+1} - 6a_n$$ for all $n$ greater than or equal to $0$.

This is the recursive sequence I have. Find a solution of the recursive sequence that satisfies the initial conditions $a_0 = 3$ and $a_1 = 11$.

Main problem is I'm not familiar enough with recursive solutions to understand what I'm being asked. If I was given another sequence and asked to show does it satisfy the original one or not I could do that, but not this. Do I generate terms? Or it's not that simple. Have a great day everyone.

Answer 1

$$a_{n+2}-2a_{n+1}=3a_{n+1}-6a_n.$$ Let $b_n=a_{n+1}-2a_n.$

Thus, $b_0=11-6=5$ and $b_{n+1}=3b_n$, which gives $b_n=5\cdot3^n$.

Hence, $a_{n+1}-2a_n=5\cdot3^n$.

Now, $$a_1-2a_0=5\cdot3^0$$ $$\frac{1}{2}a_2-a_1=\frac{1}{2}\cdot5\cdot3^1=5\cdot\left(\frac{3}{2}\right)^1$$ $$\frac{1}{2^2}a_3-\frac{1}{2}a_2=5\cdot\left(\frac{3}{2}\right)^2$$ $$.$$ $$.$$ $$.$$ $$\frac{1}{2^{n-1}}a_n-\frac{1}{2^{n-2}}a_{n-1}=5\cdot\left(\frac{3}{2}\right)^{n-1}$$ and after summing of these equalities we obtain: $$\frac{1}{2^{n-1}}a_n-2a_0=5+5\left(\frac{3}{2}\right)^1+...+5\left(\frac{3}{2}\right)^{n-1}=\frac{5\left(\left(\frac{3}{2}\right)^n-1\right)}{\frac{3}{2}-1}=10\left(\left(\frac{3}{2}\right)^n-1\right),$$ which gives $$a_n=5\cdot3^n-2^{n+1}.$$ Done!

Answer 2

Hint: $a_{n+2} - 3a_{n+1} = 2a_{n+1} - 6a_{n} $

Answer 3

The most general solution of the recurrence relation is a linear combination of those choices if $\lambda^n$ that are solutions. From $\lambda ^2=5\lambda -6$ we have $\lambda \in\{ 2,\,3\}$, so constants $a,\,b$ exist with $a_n=a2^n+b3^n$. Solving $a+b=3,\,2a+3b=11$ gives $a=-2,\,b=5$, in agreement with Michael's solution.