If $a_n = \frac{1}{2} (a_{n-1}+a_{n-2})$ prove $a_n=2+4 \left(\frac{-1}{2}\right)^n$ for $n \geq 2$

Hint:  rewrite the recursion as $2(a_n - a_{n-1})=-(a_{n-1}-a_{n-2})$. With $b_n = a_n - a_{n-1}\,$: $$b_n = -\,\frac{1}{2} \,b_{n-1}$$

Therefore $b_n$ is a geometric progression, and $a_n=b_n+a_{n-1}=b_n+b_{n-1}+a_{n-2}=\cdots$

By a well known theorem the general solution of $a_n = \dfrac{1}{2}(a_{n-1} + a_{n-2})$ can be found by means of the characteristic equation: $$\lambda^2=\dfrac{1}{2}(\lambda+1)=0\Leftrightarrow \ldots \Leftrightarrow \lambda=1\vee\lambda=-1/2,\text{ so }a_n=C_1(1)^n+C_2\left(-\dfrac{1}{2}\right)^n.$$ $$n=1\Rightarrow 0=C_1-\dfrac{1}{2}C_2,\quad n=2\Rightarrow 3=C_1+\dfrac{1}{4}C_2.$$ Solving the system we get $C_1=2,\;C_2=4$, so $$\boxed{\;a_n=2+4\left(-\dfrac{1}{2}\right)^n\;}$$