Mathematical induction proof that $f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$ 
The function $f(n)$ for $n=0,1...$ has the recursive definition $$f(n)= \begin{cases} 2 & \text {for n=0} \\ -f(n-1)+1 & \text{for n=1,2...} \end{cases}$$
  Prove by induction that the following equation holds: $$f(n)=\frac{1}{2}+\frac{3}{2}(-1)^n$$

So, I begin by checking that the basic step holds


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*$f(0)=\frac{1}{2}+\frac{3}{2}(-1)^0=2$ OK

*Assume that the equation holds for a given $n$

*Show that n+1 holds: $f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n+1} \Rightarrow f(n+1)=\frac{1}{2}+\frac{3}{2}(-1)^{n} \cdot (-1) = -f(n)-\frac{1}{2}$


I get kind of stuck here. Any advice on how I should approach this?

Well, you have come close in proving the correct form of $f(n+1)$. We then just need to do one more step. Notice that: $$f(n+1)= \frac12+ \frac32(-1)^{n+1}= (1)-(\frac12 +\frac32(-1)^n) =1-f(n)$$ proving that $f(n+1)$ is also true. The proof is thus finished!
