Suppose $A=\{a_1,a_2,\dots,a_n\}$, then
$$\mathcal P(A)=\{\emptyset,\overbrace{\{a_1\},\{a_2\},\dots,\{a_n\}}^{\binom{n}{1} \text{ sets}},\overbrace{\{a_1,a_2\},...}^{\binom{n}{2}\text{ sets}},\overbrace{\{a_1,a_2,a_3\}}^{\binom{n}{3} \text{ sets}},...,A\}.$$
Recall that $\binom{n}{k}$ tells you the number of sets you can make with $k$ elements from a set of $n$ elements.
Then if we add everything:
$$|\mathcal P(A)|=\sum\limits_{i=0}^n\binom{n}{i}=\sum\limits_{i=0}^n\binom{n}{i}(1^i)(1^{n-i})\stackrel{\text{by Binomial Theorem}}{=}(1+1)^n=2^n.$$
Why do we sum $\binom{n}{0}$ and $\binom{n}{n}$? Because of the number of sets with no elements (this is the empty set) and the number of sets with $n$ elements (this is $A$). Therefore you are interested in:
$$\sum\limits_{i=0}^2\binom{2^n}{i}.$$
Your argument is correct.
Lets see if recursion helps.
Let $[n]$ denote the set $\{1,2,\ldots,n\}$, and let $f_n$ denote the
number of minimal selfish subsets of $[n]$. Then the number of
minimal selfish subsets of $[n]$ not containing $n$ is equal to
$f_{n-1}$. On the other hand, for any minimal selfish subset of $[n]$
containing $n$, by subtracting 1 from each element, and then taking
away the element $n-1$ from the set, we obtain a minimal selfish
subset of $[n-2]$ (since $1$ and $n$ cannot both occur in a selfish
set). Conversely, any minimal selfish subset of $[n-2]$ gives rise to
a minimal selfish subset of $[n]$ containing $n$ by the inverse
procedure. Hence the number of minimal selfish subsets of $[n]$
containing $n$ is $f_{n-2}$. Thus we obtain $f_n=f_{n-1}+f_{n-2}$.
Since $f_1=f_2=1$, we have $f_n=F_n$, where $F_n$ denotes the $n$th
term of the Fibonacci sequence.
Best Answer
Hint: There is a natural bijection between your sets and the subsets with $n+1$ or more elements, via complementation. And all together, your sets and their complements make up all subsets.