Solving for $\omega_i$ given the recursive sequence $\omega_{i + 1} = \frac{7}{13}\omega_i + \frac{6}{13}\frac{v}{r}$ and $\omega_0 = 0$. I am having trouble solving this recursive sequence
$$\omega_{i + 1} = \frac{7}{13}\omega_i + \frac{6}{13}\frac{v}{r}$$
given that $\omega_0 = 0$.
I tried using Engineer's Induction  but all it's leading to are messy fractions.
 A: Well, assuming $v$ and $r$ are constants, and defining $\alpha = \frac{v}{r}$ then this is an arithmetico-geometric sequence and has the solution
$$\omega_n=\alpha\left(1-\left(\frac{7}{13}\right)^n\right)$$
EDIT: An explanation of my solution.
Let's consider a general AGP:
$$a_n=p \cdot a_{n-1} + q$$
With an initial value $a_0 = A$.
Then
$$a_0=A$$
$$a_1=pA+q$$
$$a_2=p(pA+q)+q$$
$$= p^2A + (p+1)q$$
$$a_3=p(p^2A+(p+1)q)+q$$
$$ = p^3A + (p^2+p+1)q$$
The pattern continues:
$$a_n = p^nA + q\sum_{k=0}^{n-1}{p^k}$$
Using the formula for the partial sum of a geometric series,
$$a_n = p^nA + q\frac{1-p^{n}}{1-p}$$
In OP's case, $p=\frac{7}{13}$, $q=\frac{6\alpha}{13}$, and $\omega_0 = 0$ thus
$$\omega_n = \frac{6\alpha}{13}\frac{1-\left(\frac{7}{13}\right)^n}{1-\frac{7}{13}}=\alpha\left(1-\left(\frac{7}{13}\right)^{n}\right)$$
A: A simple way to see it.
Consider
$$a_n=p \, a_{n-1} + q\qquad \qquad (p\ne 1)$$ Let $a_n=b_n+k$ and replace
$$b_n+k=p \, b_{n-1}+pk + q$$ Choose $k$ such that
$$k=pk+q \implies k=\frac q{1-p}\implies b_n=p \, b_{n-1}\implies b_n=C\, p^{n-1}$$ Back to $a$
$$a_n=C\, p^{n-1}+\frac q{1-p}\implies C=p \left(a_0-\frac{q}{1-p}\right)$$
