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

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?

• check your $a,b$ again. Does not hold when $n=1.$ – dezdichado Jan 27 '19 at 19:23
• now your $b_1$ is incorrect. – dezdichado Jan 27 '19 at 19:43
• @dezdichado how do I find $b_1$ then? I used method located in comments from this post:(link at the end) So I thought $b_1$ is the same as $b_0$ but with '$-$' math.stackexchange.com/questions/3089757/… – Gorosso Jan 27 '19 at 19:47
• $b_1 = a_1+\frac 12 = \frac 32.$ – dezdichado Jan 27 '19 at 19:52
• I think you should start with a solution that looks like $a r_1^n + b r_2^n + c$, where $r_1$ and $r_2$ are roots of your characteristic polynomial (which you have already computed). – Aditya Dua Jan 27 '19 at 20:02

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 $$a_n=b_n-\frac12$$.

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

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

• Does C1=a and C2= b? If so, I got $$b=\frac{1}{8}$$ and $$a=-\frac{1}{8}$$ If I try to calculate $a_2$ from your last line I get results=$$\frac{30}{8}$$ which is not equal to 10. – Gorosso Jan 27 '19 at 21:20
• @Gorosso Please. See note attached. – Cesareo Jan 27 '19 at 22:17
• $$C_1=-C_2+\frac{1}{2}$$ $$3C_2-\frac{3}{2}+5C_2-\frac{1}{2}=1$$ $$8C_2=3$$ $$C_2=\frac{3}{8}$$ $$\frac{1}{8}*(-3)^n+\frac{3}{8}*5^n-\frac{1}{2}$$ Am I doing something wrong here? Even if I use my version or yours $a_2$ is still not equal to 10 – Gorosso Jan 27 '19 at 22:37