Induction proof with no terms of sequence

Question

The sequence $[x_n]$ is given by $x_1=1$ and $x_{n+1}=\displaystyle\frac{4+x_n}{1+x_n}$ for $n\ge 1$. Prove by induction that for $n\ge 1$, $x_n=2\displaystyle\left(\frac{1+\alpha^n}{1-\alpha^n}\right)$ where $\alpha=\frac{-1}{3}$. I don't know how to prove this by induction when there is no $a_{k+1}$.

Answer 1

The problem:

$x_1 = 1$ and $x_{n+1} =\dfrac{4+x_n}{1+x_n} $.

Show that $x_n = 2\dfrac{1+a^n}{1-a^n} $ where $a = -\dfrac{1}{3}$.

My solution, a la Andrew D:

If $n = 1$, $2\dfrac{1+a^n}{1-a^n} =2\dfrac{1-1/3}{1+1/3} =2\dfrac{2/3}{4/3} =1 $, which matches.

For $n \ge 1$, if $x_n = 2\dfrac{1+a^n}{1-a^n} $,

$\begin{align} x_{n+1} &=\dfrac{4+x_n}{1+x_n}\\ &=\dfrac{3+1+x_n}{1+x_n}\\ &=\dfrac{3}{1+x_n}+1\\ &=\dfrac{3}{1+2\dfrac{1+a^n}{1-a^n}}+1\\ &=\dfrac{3(1-a^n)}{(1-a^n)+2(1+a^n)}+1\\ &=\dfrac{3-3a^n}{3+a^n}+1\\ &=\dfrac{3-3a^n+3+a^n}{3+a^n}\\ &=\dfrac{6-2a^n}{3+a^n}\\ &=2\dfrac{3-a^n}{3+a^n}\\ &=2\dfrac{1-a^n/3}{1+a^n/3}\\ \end{align} $

But, since $a = -\dfrac1{3}$, $\dfrac{a^n}{3} =\dfrac1{3}a^n =(-a)(a^n) =-a^{n+1} $, so $x_{n+1} =2\dfrac{1-a^n/3}{1+a^n/3} =2\dfrac{1+a^{n+1}}{1-a^{n+1}} $ which is what we want.