The given problem asks to prove, using induction, that the formula:

$\sum_{k=1}^{n} k! \cdot k = (n+1)! - 1$

holds for all positive integers $n$. Let’s break down the proof step by step using mathematical induction.

Step 1: Base Case ($n = 1$)

For $n = 1$, the left-hand side (LHS) is:

$\sum_{k=1}^{1} k! \cdot k = 1! \cdot 1 = 1$

The right-hand side (RHS) is:

$(1+1)! - 1 = 2! - 1 = 2 - 1 = 1$

So, the base case holds because LHS = RHS when $n = 1$.

Step 2: Inductive Hypothesis

Assume that the formula holds for some positive integer $n = m$, i.e.,

$\sum_{k=1}^{m} k! \cdot k = (m+1)! - 1$

Step 3: Inductive Step

We need to prove that the formula also holds for $n = m + 1$, i.e.,

$\sum_{k=1}^{m+1} k! \cdot k = (m+2)! - 1$

Using the inductive hypothesis, we can write:

$\sum_{k=1}^{m+1} k! \cdot k = \left( \sum_{k=1}^{m} k! \cdot k \right) + (m+1)! \cdot (m+1)$

From the inductive hypothesis, we know:

$\sum_{k=1}^{m} k! \cdot k = (m+1)! - 1$

Thus, the sum becomes:

$\sum_{k=1}^{m+1} k! \cdot k = (m+1)! - 1 + (m+1)! \cdot (m+1)$

Factor out $(m+1)!$:

$\sum_{k=1}^{m+1} k! \cdot k = (m+1)! \left(1 + (m+1)\right) - 1$

Simplifying:

$\sum_{k=1}^{m+1} k! \cdot k = (m+1)! \cdot (m+2) - 1 = (m+2)! - 1$

Thus, the formula holds for $n = m + 1$.

Step 4: Conclusion

By the principle of mathematical induction, the formula

$\sum_{k=1}^{n} k! \cdot k = (n+1)! - 1$

is true for all positive integers $n$.

Would you like more details or have any questions?

Related Questions:

1. What is the factorial of a number, and how is it computed?
2. Can you explain the principle of mathematical induction in detail?
3. How do we apply factorial properties to other summation problems?
4. Can this formula be generalized or modified for other functions involving factorials?
5. Are there other methods to prove this formula besides induction?

Tip:

When working with factorial expressions, always look for common factors to simplify calculations, especially in inductive proofs.