What is the nth term of this recursive sequence? This is my function:
f(x)=f(x-1)*100+1
I am curious to know whether this can be expressed as a sequence.
 A: Hint:
$$f(n + 3) = 100f(n + 2) +1 = 100(100f(n + 1) + 1) + 1 $$ $$= 100(100(100f(n) + 1) + 1) + 1= 100^3f(n) + \sum_{k = 0}^2 100^k.$$
Do you see where this is going? Once you get this, you can start with $f(0)$ and then compute $f(n) = f(n + 0)$ from the relation.
A: This is a non-homogeneous linear difference equation.
We shall solve the homogeneous equation first, then find a particular solution and finally add them to form the general solution.
Subtracting
$$f(n+1)=100f(n)+1$$
from
$$f(n+2)=100f(n+1)+1$$
we obtain the second-order homogeneous equation
$$f(n+2)-f(n+1)=100f(n+1)-100f(n)$$
or 
$$f(n+2)-101f(n+1)+100f(n)=0$$
The roots of the characteristic polynomial
$$\lambda^2-101\lambda+100$$
are 
$$\lambda_1=1$$
and
$$\lambda_2=100$$
Therefore the homogeneous solution is
$$f(n)=C\lambda_1^n+D\lambda_2^n=C+D100^n$$
As a particular solution, let us consider $f(n)$ to be a constant 
$c$
and insert into 
$$f(n+1)=100f(n)+1,$$
so
$$c=100c+1$$
and 
$$c=-\frac{1}{99}$$
Now $C$ and $D$ in the homogeneous solution are found using the initial conditions that $f(n)=f(0)$ when $n=0$ and $f(n)=100f(0)+1$ when $n=1$.
Finally, the complete solution is

$$f(n)=\dfrac{\left(99f(0)+1\right)100^n-1}{99}$$

(which may be obtained faster here)
A: $f(x) = g(x) + C$
$g(x) +C = 100g(x-1) +100C + 1$
Let $C =-\frac{1}{99}$
$g(x) = g(0) * 100^x$
$f(x) = (f(0) + \frac{1}{99}) * 100^x - \frac{1}{99}$
