Answer
The identity is false in general. It works for small $n$ but fails for $n = 5$
Work Step by Step
We are given:
- Let \( S \) be the set of all **nonempty subsets** of \( \{2, 3, \dots, n\} \)
- For each subset \( S_i \in S \), define \( P_i \) to be the **product** of the elements in \( S_i \)
- The claim is:
\[
\sum_{i=1}^{2^{n-1} - 1} P_i = \frac{(n+1)!}{2} - 1
\]
We are to **prove or disprove** this identity.
---
## ✅ Step 1: Understand the Structure
Let’s define:
\[
A = \{2, 3, \dots, n\}
\Rightarrow |A| = n - 1
\]
So:
- The number of **nonempty** subsets of \(A\) is \(2^{n-1} - 1\)
- For each such subset \(S_i\), compute the **product** of its elements: \(P_i = \prod_{x \in S_i} x\)
Then compute:
\[
\sum_{\substack{S_i \subseteq A \\ S_i \ne \emptyset}} P_i
\]
And show whether it equals:
\[
\frac{(n+1)!}{2} - 1
\]
---
## ✅ Step 2: Try Small Values of \(n\)
### Let’s test \(n = 2\):
Then \(A = \{2\}\)
- Nonempty subsets: \(\{2\}\)
- Only one subset → product = 2
Left side:
\[
\sum P_i = 2
\]
Right side:
\[
\frac{(n+1)!}{2} - 1 = \frac{3!}{2} - 1 = \frac{6}{2} - 1 = 3 - 1 = 2
\]
✅ Matches.
---
### Try \(n = 3\):
Then \(A = \{2, 3\}\)
Subsets:
- \(\{2\} \rightarrow 2\)
- \(\{3\} \rightarrow 3\)
- \(\{2,3\} \rightarrow 2 \cdot 3 = 6\)
Sum: \(2 + 3 + 6 = 11\)
Right side:
\[
\frac{(n+1)!}{2} - 1 = \frac{4!}{2} - 1 = \frac{24}{2} - 1 = 12 - 1 = 11
\]
✅ Matches.
---
### Try \(n = 4\):
Then \(A = \{2, 3, 4\}\)
All **nonempty subsets**:
- \(\{2\} \rightarrow 2\)
- \(\{3\} \rightarrow 3\)
- \(\{4\} \rightarrow 4\)
- \(\{2,3\} \rightarrow 6\)
- \(\{2,4\} \rightarrow 8\)
- \(\{3,4\} \rightarrow 12\)
- \(\{2,3,4\} \rightarrow 24\)
Sum:
\(2 + 3 + 4 + 6 + 8 + 12 + 24 = 59\)
Right side:
\[
\frac{(5)!}{2} - 1 = \frac{120}{2} - 1 = 60 - 1 = 59
\]
✅ Matches.
---
### Try \(n = 5\):
Then \(A = \{2,3,4,5\}\)
There are \(2^4 - 1 = 15\) nonempty subsets.
Let’s compute their products:
- 1-element:
\(\{2\} = 2,\; \{3\} = 3,\; \{4\} = 4,\; \{5\} = 5\)
- 2-element:
\(\{2,3\} = 6,\; \{2,4\} = 8,\; \{2,5\} = 10,\; \{3,4\} = 12,\; \{3,5\} = 15,\; \{4,5\} = 20\)
- 3-element:
\(\{2,3,4\} = 24,\; \{2,3,5\} = 30,\; \{2,4,5\} = 40,\; \{3,4,5\} = 60\)
- 4-element:
\(\{2,3,4,5\} = 120\)
Sum:
\[
2 + 3 + 4 + 5 + 6 + 8 + 10 + 12 + 15 + 20 + 24 + 30 + 40 + 60 + 120 = 379
\]
Right side:
\[
\frac{6!}{2} - 1 = \frac{720}{2} - 1 = 360 - 1 = 359
\]
❌ **Does not match**!
---
### 🚨 Conclusion:
The given identity works for \(n = 2, 3, 4\), but **fails** for \(n = 5\).
