Solve the recurrence relation $a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$, $a_0=0$, $a_1=1$

Question

My task: $a_n=2a_{n-1}+15a_{n-2}+8$ for $n\geq2$, $a_0=0$, $a_1=1$

My solution $x^{2}-2x-15$

$\Delta=64$

$x1=-3 $

$x2=5$

So I am gonna use following formula: $a_n=ar^{n}+br^{n}$

$a_n=a*(-3)^{n}+b*5^{n}$

$+8$ is the problem, so I am looking for $c$ that $b_n:=a_n+c\implies b_n=2b_{n-1}+15b_{n-2}$

$$b_n=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c$$ I am setting $c=\frac{1}{2}$ so

$$b_n=2b_{n-1}+15b_{n-2}\implies\exists a,\,b:\,b_n=a*(-3)^{n}+b*5^{n}.$$From $b_0=\frac{1}{2},\,b_1=-\frac{1}{2}$, after finding $a,\,b$. Then $a_n=b_n-\frac{1}{2}$.

$$a=\frac{3}{8}$$ $$b=\frac{1}{8}$$ $b_2=\frac{52}{8}$

$a_2=\frac{52}{8}-\frac{1}{2}=6$

Actual $a_2=10, a_3$=43

So $a_2$ from $b_n$ method is not equal to actual $a_n$.

It means I am doing something wrong here, could anyone point out the mistake?

Answer 1

Define $$b_n:=a_n+c=2(b_{n-1}-c)+15(b_{n-2}-c)+8+c=2b_{n-1}+15b_{n-2}+8-16c.$$Choosing $c=\frac12$, $$b_n=2b_{n-1}+15b_{n-2}\implies\exists a,\,b:\,b_n=a(-3)^n+b5^n.$$You an obtain $a,\,b$ from $b_0=\frac12,\,b_1=\frac{3}{2}$. Then $a_n=b_n-\frac12$.

Answer 2

Assuming $a(n) = \gamma^n$ and substituting into the homogeneous

$$ \gamma^n-2\gamma^{n-1}-15\gamma^{n-2}=0\to \gamma^n\left(1-\frac{2}{\gamma}-\frac{15}{\gamma^2}\right)=0 $$

and solving for $\gamma$ we have

$$ a_h(n) = C_1(-3)^n + C_2 5^n $$

and the particular dictates

$$ a_p(n)-2a_p(n-1)-15a_p(n-2) = 8 $$

so making $a_p(n) = C_0$ and substituting into the particular we have

$$ C_0-2C_0-15C_0 = 8\to C_0 = -\frac 12 $$

and finally

$$ a(n) = a_h(n)+a_p(n) = C_1(-3)^n + C_2 5^n-\frac 12 $$

NOTE

$$ a(0) = C_1+C_2-\frac 12 = 0\\ a(1) = C_1(-3)+C_25-\frac 12 =1 $$

and solving for $C_1, C_2$ gives

$$ a(n) = \frac 18\left(-4+(-3)^n+3 \cdot 5^n\right) $$