# Thread: f(n) general term of recurrence relation

1. ## f(n) general term of recurrence relation

I have to find the general term f(n), n>= 1, for the recurrence relation
f(n) = 3f(n-1), with f(1) = 4

I assume there is a formula or a set of rules to follow but I can not decifer it from my notes.
So I used a bit of trial and error + algebra

f(1) = 4
f(2) = 3*f(2-1) = 3*f1 = 12
f(3) = 3*f(3-1) = 3*f2 = 36
f(4) = 3*f(4-1) = 3*f3 = 108

so
$\displaystyle \frac{4}{3} * \frac{3}{1} = \frac{12}{3} = \frac{f(2)}{3}$ = f(1)
$\displaystyle \frac{36}{3} = \frac{f(3)}{3}$ = f(2)
$\displaystyle \frac{180}{3} = \frac{f(4)}{3}$ = f(3)

therefore
$\displaystyle f(n) = \frac{f(n+1)}{3}$ <-- the general term

is this ok?

2. Yes, except that usually recurrence equations are written in the form f(n + 1) = F(f(n)) for some expression F.

Edit: The above is incorrect. It is just the original equation in a different form. Apparently, you need the solution, i.e., the formula f(n) = F(n) for some expression F that consists of arithmetic operations and n but not f.

Note that

f(2) = 3 * f(1)
f(3) = 3 * f(2) = 3^2 * f(1)
f(4) = 3 * f(3) = 3 * 3^2 * f(1) = 3^3 * f(1)

You can guess what f(n) is.

3. $\displaystyle f(n) = 4\cdot 3^{n-3}$

4. f(n) = 3n-1 * f(1)
or should it be
f(n) = 3n-1 * 4

???
thanks for the help emakarov

5. This depends on the precise definition, but in my opinion, the solution (the general term) is a formula f(n) = F(n) for some expression F that consists of arithmetic operations and n, but not f.

I also suppose you understand that it's $\displaystyle 3^{n-1}$, not $\displaystyle 3n-1$.