Understanding the Statement P(n)

The given statement is \(P(n): 1 \cdot 1! + 2 \cdot 2! + 3 \cdot 3! + \dots + n \cdot n! = (n + 1)! - 1\). This statement describes a potential relationship between a sum involving factorials and a simple expression involving a factorial.

We need to determine for which values of \(n\) this statement \(P(n)\) is true. The options suggest checking different ranges of \(n\).

Verifying P(n) Using Mathematical Induction

A common method to prove statements involving sums over natural numbers is Mathematical Induction. Let's use this method to check the validity of \(P(n)\) for positive integers \(n\), starting from \(n=1\).

Base Case: Checking for n = 1

Let's substitute \(n=1\) into the statement \(P(n)\):

• Left Hand Side (LHS) = \(1 \cdot 1!\)
• \(1 \cdot 1! = 1 \cdot 1 = 1\)
• Right Hand Side (RHS) = \((1 + 1)! - 1\)
• \((1 + 1)! - 1 = 2! - 1 = 2 - 1 = 1\)

Since LHS = RHS (\(1 = 1\)), the statement \(P(1)\) is true.

Inductive Hypothesis: Assume P(k) is True

Assume that the statement \(P(n)\) is true for some positive integer \(k\), where \(k \ge 1\). That is, assume:

\[1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k! = (k + 1)! - 1\]

This is our inductive hypothesis.

Inductive Step: Proving P(k+1) is True

Now, we need to prove that the statement \(P(n)\) is also true for \(n = k + 1\). The statement \(P(k+1)\) is:

\[1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k! + (k+1) \cdot (k+1)! = ((k+1) + 1)! - 1\]

which simplifies to:

\[1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k! + (k+1) \cdot (k+1)! = (k+2)! - 1\]

Let's start with the LHS of \(P(k+1)\):

LHS = \( (1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k!) + (k+1) \cdot (k+1)! \)

Using the inductive hypothesis (the assumption that \(1 \cdot 1! + 2 \cdot 2! + \dots + k \cdot k! = (k + 1)! - 1\)), we can substitute the sum of the first \(k\) terms:

LHS = \( ((k + 1)! - 1) + (k+1) \cdot (k+1)! \)

Now, let's rearrange the terms and factor:

LHS = \( (k+1)! + (k+1) \cdot (k+1)! - 1 \)

Factor out \((k+1)!\) from the first two terms:

LHS = \( (k+1)! \cdot (1 + (k+1)) - 1 \)

Simplify the expression inside the parenthesis:

LHS = \( (k+1)! \cdot (k+2) - 1 \)

Recall the definition of factorial: \((m)! \cdot (m+1) = (m+1)!\). In our case, let \(m = k+1\).

\((k+1)! \cdot (k+2) = (k+2)!\)

So, the LHS becomes:

LHS = \( (k+2)! - 1 \)

This is equal to the RHS of \(P(k+1)\).

Conclusion from Induction and Option Analysis

We have shown that:

• The statement \(P(n)\) is true for \(n=1\) (Base Case).
• If \(P(k)\) is true for some positive integer \(k\), then \(P(k+1)\) is also true (Inductive Step).

By the principle of mathematical induction, the statement \(P(n): 1 \cdot 1! + 2 \cdot 2! + \dots + n \cdot n! = (n + 1)! - 1\) is true for all positive integers \(n\).

Let's look at the given options:

• For n > 1: This is true, but the statement is also true for n=1. So, this is not the complete set of values.
• For n > 4: This is true, but it's also true for n=1, 2, 3, 4. This is too restrictive.
• For all values of n: In the context of such mathematical statements involving sequences and sums, "all values of n" typically refers to all positive integers for which the statement is naturally defined (i.e., \(n \ge 1\)). Our induction proof shows it is true for all such \(n\).
• For all negative values of n: Factorial is not standardly defined for negative integers in this context, so the statement is not applicable or true for negative values of n.

Therefore, the statement \(P(n)\) is true for all positive integer values of \(n\). The option "For all values of n" is the most appropriate choice representing this set.