Let $S$ be a set whose elements are non-zero natural numbers. Let $f$ be a function such that $f(S)$ = square of the product of all elements in $S$. For example, if $S =\{2, 4, 5\}$ then $f(S) = (2\cdot4\cdot5)^2 = 40^2 = 1600$. Now consider $S_n = \{1, 2, 3, 4…, n\}$, the set of the first $n$ natural numbers greater than $0$. For each subset $T$ of $S_n$ that is not empty and does not contain two consecutive integers, we calculate $f(T)$, and determine its sum $Σ$. As an example, consider $S_3 = \{1, 2, 3\}$. The subsets of $S_3$ which are not empty and do not contain consecutive numbers are: $\{\{1\}, \{2\}, \{3\}, \{1, 3\}\}$. Plugging into the previously described function gives us $f({1}) = 1, f({2}) = 4, f({3}) = 9, f({1, 3}) = 9$. The summation $Σ = 1+4+9+9=23$.
Prove by mathematical induction that the summation $Σ = (n+1)!-1$.
My attempt: I believe I have established the correct base case.
Base case: n=1. Then we have $S_1=\{1\}$. Taking $f(1)$ we have $1^2=1$. So, the sum of applying $f(S)$ to all the subsets of $S1$ is $1$.
Using the given expression, $(1+1)!-1=2!-1=2-1=1$.
How would I develop the inductive step? I believe I would start by assuming the proposition holds on set $S_k$ and proceed to show it holds on set $S_{k+1}$.
