Determine a formula where $f: \mathbb{N}\rightarrow \mathbb{N} $ and f is given by $ f(n+2)=(\frac{1}{2})(f(n+1)+f(n))$ and $f(1)=1, f(2)=2$. I can't manage to transcribe a recursive formula to an explicit function.
This is not a duplicate question, as those answers address the convergence of the sequence. This is asking for a conversion into a new form.
Let $g(n)$ denote $f(n+1)-f(n)$. From the given formula, we have $$g(n+1)=-\frac{1}{2}g(n)$$ with $g(1)=1$. I think you can do the rest.
Alternatively, write $f(n)=Aa^n+Bb^n$ for some constants $A,B,a,b$. The constants $a$ and $b$ are roots of the characteristic polynomial $x^2=\frac12(x+1)$.
The question asks for a function $f\colon \mathbb{N} \to \mathbb{N}$, with $f(1) = 1$ and $f(2) = 2$ and $f(n+2) = (f(n+1) + f(n))/2$.
No such function exists, as $f(n+2) = (f(n+1)+f(n))/2$ would imply that $f(3) = (f(2)+f(1))/2 = (2+1)/2 = 3/2$, and that is not a natural number.
