Recursive Sequence solving for $f(200)$ Let $f$ be defined recursively by: $f(0)=5$ and $f(n+1)=3f(n)-2$. Find $f(200)$
I'm really confused how to go about solving this. Can someone help? Thank you! 
 A: Hint: $f_{n+1} - 1 = 3\left(f_n - 1\right)$. Does this help?
A: You could think of the first 5 terms as 
$x, \space 3x-2,  \space 9x-8,  \space 27x-26,  \space 81x-80$ where $x=5$ and infer what the equation for the 200th term would be.
ETA: you could think of the terms as $(5), \space 3(5)-2,  \space 9(5)-8,  \space 27(5)-26,  \space 81(5)-80$.  Don't need to add another variable.
A: Expanding on
OC-Sansoo's answer:
Since
$f(n+1)=3f(n)-2$,
subtracting one from each side,
$f(n+1)-1
=3f(n)-3
=3(f(n)-1)
$.
Letting
$g(n)
=f(n)-1
$,
$g(n+1)
=3g(n)
$.
By induction,
$g(n)
=3^n g(0)
$
or
$f(n)-1
=3^n(f(0)-1)
=4\cdot 3^n
$
or
$f(n)
=4\cdot3^n+1
$.
In general,
if
$f(n+1)
=af(n)+b
$,
we want to find $c$
such that,
if $g(n) = f(n)+c$,
then
$g(n+1) = ag(n)$
(so that
$g(n) = a^n g(0)$).
Since $f(n) = g(n)-c$,
$g(n+1)-c
=a(g(n)-c)+b
=ag(n)-ac+b
$
or
$g(n+1)
=ag(n)+c-ac+b
=ag(n)+c(1-a)+b
$.
If we choose
$c(1-a)+b=0$,
or
$c
=\frac{b}{a-1}
$,
then
$g(n) = a^n g(0)$
or
$f(n)+c = a^n(f(0)+c)$
or
$f(n) 
= a^nf(0)+c(a^n-1)
= a^nf(0)+\frac{b(a^n-1)}{a-1}
$.
Note that does not work if $a=1$,
but then
$f(n)$ is linearly i  ncreasing,
not growing by a power law.
