Question

Use mathematical induction to show that 2^n > n^2 + n whenever n is an integer greater than 4.

Answer 1

To prove 2^n > n^2 + n for integers n > 4 via mathematical induction, we check the base case at n=5, which is true. We then assume the statement for k and prove for k+1, showing that 2^(k+1) > (k+1)^2 + (k+1), thus confirming the inequality by induction.

Step-by-step explanation:

To use mathematical induction to demonstrate that 2n > n2 + n for all integers n > 4, we perform the following two steps:

1. Base Case: Verify the statement for the initial integer, which is n=5 in this case.
2. Inductive Step: Assume the statement holds for some integer k > 4 (inductive hypothesis) and then prove it for k+1.

For the base case, when n=5:

25 = 32 > 52 + 5 = 25 + 5 = 30, which is true.

For the inductive step, assume 2k > k2 + k is true for some k > 4. Now for k+1:

2k+1 = 2k * 2 > (k2 + k) * 2 = 2k2 + 2k

For 2k+1 to be greater than (k+1)2 + (k+1), it needs to be shown that:

2k2 + 2k > k2 + 2k2 + k + 1

Subtracting k2 + 2k from both sides gives:

k2 > k + 1, which is true for k > 4

Therefore, by mathematical induction, 2n > n2 + n for all integers n > 4.