Assume true for n, i.e.

$\begin{align*}&a_{n+1}-a_n=\left(-\dfrac 1 2\right)^n (a_1-a_0)\\ \\

&\text{Now show true for }n+1 \\ \\

&a_{n+2}-a_{n+1} = \dfrac 1 2 (a_{n+1} + a_n) - a_{n+1}= \\ \\

&-\dfrac 1 2 (a_{n+1}-a_n)=\\ \\

&-\dfrac 1 2 \left(-\dfrac 1 2\right)^n (a_1-a_0)= \\ \\

&\left(-\dfrac 1 2\right)^{n+1} (a_1-a_0)\end{align*}$

and as you showed it true for $n=2$ this completes the proof.