2^n<(n+1) for n>=0

Answer 1

To prove the inequality 2^n < (n+1) for n≥0, we can use mathematical induction.

Base case:
For n=0, we have 2^0 = 1, and (0+1) = 1. Since 1 < 1, the inequality holds for the base case.

Inductive step:
Assume that the inequality holds for some positive integer k, i.e., 2^k < (k+1).

Now, we need to show that the inequality holds for k+1, i.e., 2^(k+1) < (k+1+1).

Starting with 2^k < (k+1) (inductive hypothesis), we can multiply both sides by 2 to get:
2 * 2^k < 2 * (k+1)

Simplifying, we have:
2^(k+1) < 2k + 2

Since k≥0, we know that 2k + 2 > k + 2.

Therefore, 2^(k+1) < k + 2.

Since (k+1) > k, we have:
2^(k+1) < (k+1) + 1

This shows that the inequality holds for n = k+1.

By mathematical induction, we have proven that 2^n < (n+1) for n≥0.

Answer 2

what???!!!
I think you mean 2^n > n+1 for n >1
consider 2^2 > 3

Answer 3

Apologies for the confusion in my previous response. You are correct, the correct inequality to show is 2^n > n+1 for n > 1.

To prove the inequality 2^n > n+1 for n > 1, we can again use mathematical induction.

Base case:
For n = 2, we have 2^2 = 4, and 2+1 = 3. Since 4 > 3, the inequality holds for the base case.

Inductive step:
Assume that the inequality holds for some positive integer k, i.e., 2^k > k+1.

Now, we need to show that the inequality holds for k+1, i.e., 2^(k+1) > (k+1)+1.

Starting with 2^k > k+1 (inductive hypothesis), we can multiply both sides by 2 to get:
2 * 2^k > 2 * (k + 1)

Simplifying, we have:
2^(k+1) > 2k + 2

Since k > 1, we know that 2k + 2 > k + 2.

Therefore, 2^(k+1) > k + 2.

Since (k+1) > k, we have:
2^(k+1) > (k+1) + 1

This shows that the inequality holds for n = k+1.

By mathematical induction, we have proven that 2^n > n+1 for n > 1.