The questions within are regarding the following answer on this post my questions are under the link. I apologize for making you go to the link to view the question. I am still figuring this part of stackexchange out. I am mostly on stackoverflow.
https://math.stackexchange.com/a/1291504/370523
why is the base case both 1 and 2?
clearly F(1) = 1 < 2^1 = 2 and f(2) = 1 < 2^2 = 4
but f(n-2) + f(n-1) when n = 1 is -1 how is it 1?
more to be added once people start responding, I am not sure if you guys are going to rage out on me for asking these questions. So I am testing.
We need base cases of both $n=1$ and $n=2$ because for $n\geq 3$ we are going to show that $$\bullet \quad [\forall j<n \;(F(j)<n 2^j)\;]\implies [F(n)<2^n].$$ We need at least 2 successful cases (case $n-1$ and case $n-2$) to apply the formula $F(n)=F(n-1)+F(n-2$). Note that the case $n=3$ is $$[\forall j<3\;(F(j)<2^j)]\implies [F(3)<2^3]$$ which is equivalent to $$[F(1)<2^1\land F(2)<2^2]\implies [F(3)<2^3].$$ Which is done inductively as $$F(1)<2^1\land F(2)<2^2\implies F(3)=F(2)+F(1)<2^2+2^1<2^3.$$