https://file.aiquickdraw.com/mathbot/file/1729082383743ncciwr62.jpg

Let's solve each part step by step.

### Given:
The inequality is \( P(n) \), defined as:
\[
2^n < (n+1)!
\]
for every integer \( n \geq 2 \).

### Part (a):
We need to find \( P(2) \):
\[
P(2): \quad 2^2 < (2+1)!
\]
Calculating each side:
- \( 2^2 = 4 \)
- \( (2+1)! = 3! = 6 \)

So, \( P(2) \) is:
\[
4 < 6
\]
Thus, the correct choice is:
- \( 4 < 6 \)

And \( P(2) \) is **true**.

### Part (b):
We need to determine \( P(k) \), where \( P(k) \) is:
\[
2^k < (k+1)!
\]
The correct option corresponding to this expression is:
- \( 2^k < (k+1)! \)

### Part (c):
We need to find \( P(k+1) \):
\[
P(k+1): \quad 2^{k+1} < ((k+1) + 1)!
\]
This simplifies to:
\[
2^{k+1} < (k+2)!
\]
The corresponding choice is:
- \( 2^{k+1} < (k+2)! \)

### Part (d):
For a proof by mathematical induction, the inductive step requires showing that if \( P(k) \) is true, then \( P(k+1) \) is also true. Thus, the correct choice is:
- We need to show that if \( k \) is any integer with \( k \geq 2 \) and if \( P(k) \) is true, then \( P(k+1) \) is also true.

Would you like further details or have any questions about this solution?

### Related Questions:
1. How does mathematical induction work for inequalities?
2. What are some other ways to prove the inequality \( 2^n < (n+1)! \)?
3. What happens if we try this inequality for values of \( n < 2 \)?
4. Can we generalize this approach to other similar factorial-based inequalities?
5. How can we visualize the growth of \( 2^n \) and \( (n+1)! \) for large \( n \)?

### Tip:
In mathematical induction, always verify the base case before proceeding to the inductive step to ensure the validity of the proof.

For every integer n ≥ 2, let P(n) be the following inequality. 2^n < (n + 1)!

Explore how to prove the inequality 2^n < (n + 1)! for n ≥ 2 through mathematical induction. Learn key concepts involving factorial growth versus exponential growth, with step-by-step breakdowns for the base case and inductive step.

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