I have this statement:
A function $h$, whose domain is the natural numbers, verifies the property: $h(n) = \frac{h(n+1)}{2}$, $\forall n \in Dom_h$ If $h(1) = 3$, then $h(20)= $
My development was:
$h(n) = \frac{h(n+1)}{2} = \frac{h(n)+h(1)}{2}$, thus:
$2h(n) = h(n) +3 ==> h(n) = 3$, so for any value of $n$, it will be always $3$, but according to the guide my result is incorrect.
What have I done wrong? And what is the fastest way to get to the solution?
It isn't given that $\color{blue}{h(a+b) = h(a)+h(b)}$. So we can't directly write like that.
$\color{green}{h(n+1) = 2h(n) = 2\cdot2 h(n-1) = 2\cdot2\cdot2h(n-2) = \cdots 2\cdot2\cdots2h(1) =}\color{red}{2^nh(1)}$
So if $n+1 = 20 \implies n = 19$
$$h(20) = 2^{19}h(1) = 2^{19}\cdot3$$