Solving for $\omega_i$ given the recursive sequence $\omega_{i + 1} = \frac{7}{13}\omega_i + \frac{6}{13}\frac{v}{r}$ and $\omega_0 = 0$.

Question

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.

Answer 1

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 $$\omega_n = \frac{6\alpha}{13}\frac{1-\left(\frac{7}{13}\right)^n}{1-\frac{7}{13}}=\alpha\left(1-\left(\frac{7}{13}\right)^{n}\right)$$

Answer 2

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 $a$ $$a_n=C\, p^{n-1}+\frac q{1-p}\implies C=p \left(a_0-\frac{q}{1-p}\right)$$