Can you check my answers?

1.Which of the following shows the best next step to prove the following by mathematical induction?

3^n>n*2^n, n≥1
1.When n=1, the formula is valid because

3^1 1*2^1
3>2
2.Assuming that
3^k>k*2^k

a.3^k+1>k+1*2^k+1
b.3^k+1>k*2^k+1
c.4^k>k*3^k
d.3^k+1>(k+1)2^k+1

answer: A

2.Which of the following shows the best next step to prove the following by mathematical induction?

n!>2^n, n≥4

1. When n=1 the formula is valid because

4!>2^4
24>16
2.Assuming that k!>2^k
show that (k+1)!>2^k+1
(k+1)!=k!(k+1)

a.k!(k+1)>k!(k+k)
b.k!(k+1)>2^k(k+1)
c.k+1≥5
d.(k+k)>(k+1)!

answer:a

To check the answers, let's go through the steps of each question and see if the given next steps align with the process of proof by mathematical induction.

Question 1:
The statement to prove is 3^n > n*2^n for n ≥ 1.

1. The base case is n = 1:
- 3^1 > 1 * 2^1
- 3 > 2
This step checks out and is correct.

2. Assuming the formula holds for some k:
- Assuming 3^k > k * 2^k, we need to prove the same inequality for k + 1.

The options for the next step are:
a. 3^(k+1) > (k+1) * 2^(k+1)
b. 3^(k+1) > k * 2^(k+1)
c. 4^k > k * 3^k
d. 3^(k+1) > (k+1) * 2^(k+1)

The correct answer is option a because it follows the correct progression of the proof. We need to show that if the inequality holds for k, it also holds for k + 1.

Answer: A (3^(k+1) > (k+1) * 2^(k+1))

Question 2:
The statement to prove is n! > 2^n for n ≥ 4.

1. The base case is n = 4:
- 4! > 2^4
- 24 > 16
This step checks out and is correct.

2. Assuming the formula holds for some k:
- Assuming k! > 2^k, we need to prove the same inequality for k + 1.

The options for the next step are:
a. k!(k+1) > k!(k+k)
b. k!(k+1) > 2^k(k+1)
c. k+1 ≥ 5
d. (k+k) > (k+1)!

The correct answer is option a because it follows the correct progression of the proof. We need to show that if the inequality holds for k, it also holds for k + 1.

Answer: A (k!(k+1) > k!(k+k))

Based on the analysis, it seems that both answers provided are correct.