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.