prove: $\dfrac{2^{n+1}+(-1)^n}{3}$ I am asked to prove this notation with induction for $n\in \mathbb{N}$:
real problem is to fill the area with tilings. and for $n\in \mathbb{N}$ there are exactly so many chances to fill the area as the formula says:  $$\dfrac{2^{n+1}+(-1)^n}{3}$$
I started in this way:
Base Case: $n=1$: 
$$\dfrac{ 2^2-1}{3} = 1$$
Inductive step: $n\rightarrow n+1$: 
$$\dfrac{2^{n+2}+(-1)^{n+1}}{3} = help..$$
my problem is: what am i doing, i seem to have already proven it in this last step which is obviously wrong. what am i missing here in this proof?
thanks for help 

$$\dfrac{2^{n+2}+(-1)^{n+1}}{3}$$
$$\dfrac{2^{n+1}\cdot2+(-1)^{n}\cdot(-1)}{3}$$
$$\dfrac{2^{n+1}\cdot2-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3-1)-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3)-2^{n+1}-(-1)^{n}}{3}$$
$$\dfrac{2^{n+1}\cdot(3)-(2^{n+1}+(-1)^{n})}{3}$$
$$\dfrac{2^{n+1}\cdot(3)}{3}-\dfrac{(2^{n+1}+(-1)^{n})}{3}$$
Proven.
