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.

Hint: Subtract $a_{n-1}$ from both sides. We get
$$a_n-a_{n-1}=-\frac{1}{2}(a_{n-1}-a_{n-2}).$$
The "new difference" is the "old difference" times $-\frac{1}{2}$.
