Use induction to show that $a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) .$ 
Let $a_0$ and $a_1$ be distinct real numbers. Define
  $a_n=\frac{a_{n-1}+a_{n-2}}{2}$ for each positive integer $n\geq 2$. Prove that $$a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0) $$

This is what I have so far:

Let $P(n)$ be the statement: For all $n\geq 2$, $n\in\mathbb{N}$, $$
 a_{n+1}-a_n=\biggl(-\frac{1}{2} \biggr)^n (a_1-a_0). $$ We must show
   1. $P(2)$ holds, 2. For all $k\geq 2$, $k\in\mathbb{N}$, if $P(k)$ holds, then $P(k+1)$ holds. Let $n=2$. Then   $$  \begin{aligned}
 a_{n+1}-a_n &=a_3-a_2\\  &=\frac{a_2+a_1}{2}-\frac{a_1+a_0}{2}\\
 &=\frac{a_1+a_0+2a_1}{4}-\frac{a_1+a_0}{2}\\
 &=\frac{3a_1+a_0}{4}+\frac{-2a_1-2a_0}{4}\\  &=\frac{a_1-a_0}{4}\\
 &=\biggl(-\frac{1}{2}\biggr)^2(a_1-a_0).  \end{aligned}  $$    Hence,
  $P(2)$ holds.   Suppose that for all $k\geq 2$, $k\in\mathbb{N}$,
  $P(k)$ holds, that is   $$  \begin{aligned}  a_{k+1}-a_k
 &=\frac{a_k+a_{k-1}}{2}-\frac{a_{k-1}+a_{k-2}}{2}\\ 
  &=\biggl(-\frac{1}{2} \biggr)^k (a_1-a_0).  \end{aligned}  $$    Then
      $$  \begin{aligned}  a_{(k+1)+1}-a_{k+1} &=a_{k+2}-a_{k+1}\\
  &=\frac{a_{k+1}+a_k}{2}-\frac{a_k+a_{k-1}}{2}\\
  &=\frac{a_k+a_{k-1}+a_{k-1}+a_{k-2}}{4}-\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}}{4}\\  &=\biggl(\frac{a_k+a_{k-1}-a_{k-2}-a_{k-3}}{2}\biggr)^2\\
 &=\frac{a_{k-1}+a_{k-2}+a_{k-2}+a_{k-3}-a_{k-3}-a_{k-4}-a_{k-4}-a_{k-5}}{8}\\
 &=\biggl(\frac{a_{k-1}+2a_{k-2}-2a_{k-4}-a_{k-5}}{2}\biggr)^3
 \end{aligned}  $$

Ok, so here it seems that I will just end up substituting forever. I'm sure that I'm missing something obvious. Any help would be greatly appreciated. Thanks.

You should use the induction hypothesis (with two base cases) immediately after your second equality when computing $a_{k + 1} - a_{k + 1}$, for we then have
\begin{align*}
\frac{a_{k + 1} + a_k}{2} - \frac{a_k + a_{k - 1}}{2} &= \frac 1 2\left(- \frac 1 2\right)^k (a_1 - a_0) - \frac 1 2\left(-\frac 1 2\right)^{k - 1} (a_1 - a_0) \\
&= (-1)^k \left(\frac{1}{2^{k + 1}} - \frac 1 {2^k}\right) (a_1 - a_0) \\
&= (-1)^k \left(-\frac{1}{2^{k + 1}}\right) (a_1 - a_0) \\
&= \left(-\frac 1 2 \right)^{k + 1} (a_1 - a_0)
\end{align*}
as desired. 
