Crazy Jumping Frog

Solution for Question 1 The expected distance in feet for making $1$ jump is $$1\cdot p + 2\cdot (1-p)=2-p$$ Since each jump is independent, therefore, the expect distance after the frog making $n$ jumps is $n(2-p)$ feet.

Solution for Question 2 Denote the probability of the frog jumping $n$ feet is $f(n)$, we have $$f(n)=p\cdot f(n-1) + (1-p)\cdot f(n-2)\tag{1}$$ with initial values as $$f(0)=1$$ $$f(1)=p$$ $$f(2)=p\cdot f(1) + (1-p)\cdot f(0)=p^2-p+1$$
To solve (1), which is a Recurrence Relation, we need to find the roots of its characteristic equation as the following: $$x^2=px+(1-p)$$ Solve the above equation, we have $$x_{1}=1, x_{2}=p-1$$ Therefore $$f(n)=c_{1}x_{1}^n+c_{2}x_{2}^n=c_{1}+c_{2}(p-1)^n$$ Because $f(0)=1$, $f(1)=p$, we have $$ \begin{array}{lcc@} c_{1} + c_{2} & = & 1\\ c_{1} + c_{2}\cdot (p-1) & = & p \end{array} $$ Solve the above linear equations, we have $$c_{1}=\frac{1}{2-p}, \ \ \ \ c_{2}=\frac{1-p}{2-p}$$ Therefore $$f(n)=c_{1}+c_{2}(p-1)^n=\frac{1}{2-p}+\frac{1-p}{2-p}(p-1)^n=\frac{1-(p-1)^{n+1}}{2-p}$$