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.