Find the sequence $\{c_n\}$ for $c_n = \alpha \cdot c_{n-1} + {\alpha}^{\beta-n}$

Question

Let $\alpha$ and $\beta$ be two given constants, how to find the sequence $\{c_n\}$ for

$c_n = \alpha \cdot c_{n-1} + {\alpha}^{\beta-n}$,

where $c_0 = {\alpha}^{\beta}$.

Answer 1

Given the sequence, for $n \geq 1$, \begin{align} c_{n} = a c_{n-1} + a^{b-n} \end{align} then it is seen that the first few values are \begin{align} c_{1} &= a c_{0} + a^{b-1} \\ c_{2} &= a c_{1} + a^{b-2} \\ c_{3} &= a c_{2} + a^{b-3}. \end{align} Using the values accordingly \begin{align} c_{2} &= a^{2} c_{0} + a^{b-2}(a^{2} + 1) \\ c_{3} &= a^{3} c_{0} + a^{b-3}(a^{4} + a^{2} + 1). \end{align} Carrying the process further it will be seen that \begin{align} c_{n} = a^{n} c_{0} + a^{b-n} \ \frac{1-a^{2n}}{1-a^{2}}. \end{align}

Given the initial condition $c_{0} = a^{b}$ the general form becomes \begin{align} c_{n} = a^{b+n} + a^{b-n} \ \frac{1-a^{2n}}{1-a^{2}} \end{align} or \begin{align} c_{n} = a^{b-n} \ \frac{1-a^{2n+2}}{1-a^{2}}. \end{align}

Answer 2

We can prove by induction that $c_n = \sum_{k = 0}^n \alpha ^{\beta+n-2k}$.

The base cases are trivial. Now

$$c_{n+1} = \alpha c_n+\alpha^{\beta-(n+1)} = \alpha \sum_{k=0}^n \alpha^{\beta+n-2k}+\alpha^{\beta-(n+1)} = \sum_{k = 0}^n \alpha^{\beta+ (n+1)-2k}+\alpha^{\beta-(n+1)}$$

$$ = \sum_{k = 0}^{n+1} \alpha^{\beta+ (n+1)-2k}$$

which completes the induction.

Then

$$c_n = \alpha^{\beta} \left( \alpha^{-n} + \alpha^{-n+2} + \cdots + \alpha^n \right) = \alpha^{\beta} \cdot \frac{a^{2n+2}-1}{a^{n+2}-1}.$$