Solution
= Solution
{c}
The <characteristic equation of a linear recurrence> is
$$
16r^2-8r-3=(4r-3)(4r+1)=0.
$$
Its distinct roots are $3/4$ and $-1/4$, so $p_n=A(3/4)^n+B(-1/4)^n$. The <initial conditions> are $p_0=1$ and $p_1=1/2$, giving $A+B=1$ and $3A-B=2$. Hence $A=3/4$, $B=1/4$, and
$$
\boxed{p_n=\frac34\left(\frac34\right)^n+\frac14\left(-\frac14\right)^n
=\frac{3^{n+1}+(-1)^n}{4^{n+1}},\qquad n\ge0.}
$$