Please see below for my proof involving the Stirling numbers of the second kind involving $n$ distinct elements and $3$ non-empty subsets.
Looking for constructive criticism and feedback. In particular, is the simplification in the inductive step the best way forward?
Prove by induction that $$S(n,3)=\frac{3^{n-1}+1}{2}-2^{n-1}.$$
$Proof.$ We will prove the above proposition via mathematical induction.
Base Case. For Stirling numbers of the second kind we know that $S(n,n)=1$. Then let $n=3$ and observe that $$\begin{align*}S(3,3)&=\frac{3^{3-1}+1}{2}-2^{3-1}\\&=\frac{3^2+1}{2}-2^2\\&=5-4=1.\end{align*}$$
Thus for $n=3$ our equation holds.
Inductive Step. Let's assume $n=k\ge 3$ and $S(k,3)=\frac{3^{k-1}+1}{2}-2^{k-1}$. Then by the recurrence relation of Stirling numbers of the second kind observe that $$\begin{align*}S(k+1,3)&=3S(k,3)+S(k,2)\\&=3\left(\frac{3^{k-1}+1}{2}-2^{k-1}\right)+(2^{k-1}-1) \\&=\frac{3^k+3}{2}-3(2^{k-1})+2^{k-1}-1 \\&=\frac{3^k+3}{2}-1-2(2^{k-1})\\&=\frac{3^k+1}{2}-2^k.\end{align*}$$
It follows by mathematical induction that $S(n,3)=\frac{3^{n-1}+1}{2}-2^{n-1}$ for all $n\ge 3$. $\Box$
