The task is to find another relation for the adjusted Fibonacci numbers. I've found there genertaing function $$A(x)=\dfrac{1}{1-x-x^{2}}$$
Furthermore I've created the generating function in a different way and now want to grab the correct coefficient.
Since $\hat{F}_{n}$ is the number of ways to add one and two to get $n$,
$$A(x)=\sum_{n\geq 0}(x+x^{2})^{n}$$
Now I'm having trouble grabbing the $n^{th}$ coefficient.
It should be $\sum_{k=0}^{n}\binom{k}{n-k}$.
All I have so far is
$$A(x)=\sum_{n\geq 0}\sum_{k\geq 0}^{n} \binom{n}{k}x^{n-k}x^{2k}$$
Then I arrive at $$A(x)=\sum_{n\geq 0}(1+x)x^{n}$$ Which does me no good. A hint would be very nice. Thank you.
Hint: We factor $1-x-x^2=(1-\phi x)(1-\psi x)$ for appropriate real numbers $\phi,\psi$. Then solve the partial fractions problem $$\frac{1}{1-x-x^2}=\frac{A}{1-\phi x} + \frac{B}{1-\psi x}$$
Then you have the sum of two geometric series.