https://file.aiquickdraw.com/mathbot/file/1729029034935zpaykotj.jpg

The equation provided in the image is as follows:

\[
\sum_{i=1}^{n} i \cdot i! = (n+1)! - 1
\]

### Explanation:
- The sum notation on the left-hand side sums the product of \( i \) and \( i! \) (i factorial) from \( i = 1 \) to \( n \).
- The right-hand side is \( (n+1)! - 1 \), which involves the factorial of \( n+1 \), then subtracting 1 from it.

### Proof:
This identity can be proven using induction or by manipulating the factorials and sums. Let's sketch the induction proof:

1. **Base Case:**
   For \( n = 1 \),
   \[
   \sum_{i=1}^{1} i \cdot i! = 1 \cdot 1! = 1
   \]
   And on the right-hand side:
   \[
   (1+1)! - 1 = 2! - 1 = 2 - 1 = 1
   \]
   So, the base case holds.

2. **Inductive Step:**
   Assume the identity holds for \( n = k \):
   \[
   \sum_{i=1}^{k} i \cdot i! = (k+1)! - 1
   \]
   Now, for \( n = k+1 \), consider:
   \[
   \sum_{i=1}^{k+1} i \cdot i! = \left( \sum_{i=1}^{k} i \cdot i! \right) + (k+1) \cdot (k+1)!
   \]
   Using the inductive hypothesis:
   \[
   = (k+1)! - 1 + (k+1) \cdot (k+1)!
   \]
   Factor \( (k+1)! \) out:
   \[
   = (k+1)! \left(1 + (k+1)\right) - 1 = (k+2) \cdot (k+1)! - 1 = (k+2)! - 1
   \]
   Thus, the identity holds for \( n = k+1 \), completing the proof by induction.

Would you like more details on this proof or to explore alternative ways to approach it?

### Follow-up questions:
1. Can you prove this using a combinatorial interpretation of factorials?
2. How would this formula change if the factorial terms were squared?
3. Can you generalize this identity to include higher powers of \( i \) (like \( i^2 \cdot i! \))?
4. How does this identity relate to the standard sum of factorials formula \( \sum_{i=1}^{n} i! \)?
5. Could this result be derived using generating functions?

#### Tip:
When using induction, always ensure your base case is correct and the inductive step handles all cases smoothly.

For all n ∈ ℤ+, show that the sum of i * i! from i=1 to n equals (n+1)! - 1.

Explore how to prove the identity ∑(i=1 to n) i * i! = (n+1)! - 1 using mathematical induction. This problem involves factorials, summations, and the application of induction, suitable for students in Grades 11-12.

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