Still a novice could use a little help Ok, so I have this question, and we never went over this or how to solve it in class. I can't find an example in the book either.
How do I show that $f_n = 3^nA + 2^nB$ satisfies the recurrence equation $f_n = 5f_{n-1} - 6f_{n-2}$ for $n\geq2$?
What exactly is $A$ and $B$, and how do I even get started? What would be the first thing I look for? (If this is a calculus type problem, keep in mind that my highest math attained so far is trig, so sorry for being ignorant, just haven't seen this type of problem in book, power point, notes, or lecture).
 A: substituting we get:  $$5f_{n-1}-6f_{n-2}=$$
$$5(A*3^{n-1}+B*2^{n-1})-6(A*3^{n-2}+B*2^{n-2})=$$
$$5A*3^{n-1}+5B*2^{n-1}-6A*3^{n-2}-6B*2^{n-2}=$$
$$5A*3^{n-1}+5B*2^{n-1}-2A*3^{n-1}-3B*2^{n-1}=$$
$$ 3A*3^{n-1}+2B*2^{n-1}=A*3^n + B*2^n$$
In conclusion all you had to do was substitute and use laws of exponents.
In general whenever you have to prove a recurrence relation you should just evaluate for $f_n$, and then evaluate the thing they are telling you it should be equal too. If they are the same then it holds. Hope this helps.
Regards.
A: It looks like $A$ and $B$ are supposed to be numbers.  Depending on your choice of $A$ and $B$, you define a different sequence $\{f_1,f_2,\dots\}$, but every such sequence should satisfy the recurrence equation.
We verify that the recurrence relation is satisfied as follows:
$$
\begin{align}
5f_{n-1} - 6f_{n-2} &=
5(3^{n-1}A + 2^{n-1}B) - 6(3^{n-2}A + 2^{n-2}B) \\
&= (5)3^{n-1}A + (5)2^{n-1}B - (6)3^{n-2}A - (6)2^{n-2}B)\\
&= [(5)3^{n-1} - (6)3^{n-2}]A + [(5)2^{n-1} - (6)2^{n-2}]B\\
&= [(5)(3) - (6)]3^{n-2}A + [(5)(2) - 6]2^{n-2}B\\
&= (9)3^{n-2}A + (4)2^{n-2}B \\
&= 3^nA + 2^nB = f_n
\end{align}
$$
As required.
A: $f$ is a list of numbers. $f_n$ refers to the $n$-th number in the list, for instance, $f_1$ is the first number, $f_3$ is the third number, and so on. This is called a sequence.
You've been given the definition for the sequence
$$f_n=A\ 3^n+B\ 2^n$$
Meaning that the $n$-th number of the sequence is equal to the the result of the calculation on the right (you just plug in $n$). For instance, the first value of the list is $3A+2B$, the second value is $9A+4B$, and so on.
$A$ and $B$ are of course just numbers, and you can see that you're going to get a different list depending on what exactly $A$ and $B$ are. The reason the book doesn't give specific values for $A$ and $B$ is because you're supposed to show that the answer to this specific question is the same no matter what they are.
A recurrence relation for a sequence is where each value of the sequence can be calculated from previous values. For example, I could have a sequence called $S$ where each value is twice the previous value, in other words, $S_{n+1}=2S_n$. You're being asked to show that the recurrence relation:
$$f_n=5f_{n-1}-6f_{n-2}$$
Is true for the sequence $f$.
Well, they've given you the exact formula to calculate any given value of $f$. So just use that to calculate $f_n$, $f_{n-1}$ and $f_{n-2}$, and then check that the above equation is true. See the other answers for details on that part.
