Use the substitution $A_n=r^n$ to solve each of the recurrence relation

Question

Use the substitution $A_n=r^n$ to solve each of the recurrence relation:

(a)$2A_n=7A_{n-1}-3A_{n-2}; n \ge 0;A_0=A_1=1$

(b)$2S_n=7S_{n-1}-3S_{n-2}; n \ge 0;S_0=S_1=1$

This is a problem in my study guide and (a) and (b) seems the same problem to me. We can solve (a) by letting $A=r^n$ and we can solve (b) by letting $S=t^n$ or something. Both (a) & (b) should have the same solution. I am a bit perplexed why the professor will put the same problem twice. Am I missing something?

99% chance it's a typo. But if you're "perplexed why the professor will put the same problem twice," why don't you ask the professor? — , Mar 16 '18 at 21:19
n ≥ 0 That should rather be $,n \ge \color{red}{2},$. Other than that, yes, it's the same problem. — dxiv, Mar 16 '18 at 21:19
Yes ineed they seems the same problem, that's good you can solve only one! — user, Mar 16 '18 at 21:24
The professor who set up the study guide is on sabbatical and he will return next month. So, have to wait till then to ask him the question. Meanwhile, I need to study for my comprehensive exam, so reaching out here to see if anyone can help. Thanks! — nova_star, Mar 16 '18 at 21:29

Dr. Sonnhard Graubner · Accepted Answer · 2018-03-16T21:37:06.080

1

HINT: with your Substitution we get $$2q^n=7q^{n-1}-3q^{n-2}$$ this gives the solution $$a_n=\frac{1}{5}2^{-n}(4+6^n)$$ which is the same as your result.

edited Mar 16 '18 at 21:37

answered Mar 16 '18 at 21:26

Dr. Sonnhard Graubner

95,283

Solving for it, I got $A_n=1/5(2^{2-n}+3^n)$ – nova_star Mar 16 '18 at 21:32

Use the substitution $A_n=r^n$ to solve each of the recurrence relation

1 Answers1