Question

Proving inequalities by induction. Prove each of the following statements using mathematical induction. (a) Prove that for n ? 2, 3^n > 2^n + n^2. Solution Proof. By induction on n. Base case: Show that for n = 2, 3^n > 2^n + n^2: 3^2 = 9 > 8 = 2^2 + 2^2, so the inequality holds for n = 2. Inductive step: Assume that k is a positive integer and 3^k > 2^k + k^2. Show that 3^(k+1) > 2^(k+1) + (k+1)^2. 3^(k+1) = 3 ? 3^k > 3(2^k + k^2) by the inductive hypothesis = 3 ? 2^k + 3k^2 = 2 ? 2^k + 2^k + k^2 + 2k^2 = 2^(k+1) + 2^k + k^2 + 2k^2 ? 2^(k+1) + 1 + k^2 + 2k = 2^(k+1) + (k+1)^2 The second-to-last step makes use of the fact that for k ? 1, 2^k ? 1 and k^2 ? k. Therefore 3^(k+1) > 2^(k+1) + (k+1)^2. (b) For any n ? 1, the factorial function, denoted by n!, is the product of all the positive integers through n: n! = 1 ? 2 ? 3 ? ? ? (n - 1) ? n Prove that for n ? 4, n! ? 2^n.

          Proving inequalities by induction.

Prove each of the following statements using mathematical induction.

(a) Prove that for n ? 2, 3^n > 2^n + n^2.

Solution
Proof.
By induction on n.

Base case: Show that for n = 2, 3^n > 2^n + n^2:
3^2 = 9 > 8 = 2^2 + 2^2, so the inequality holds for n = 2.

Inductive step: Assume that k is a positive integer and 3^k > 2^k + k^2. Show that 3^(k+1) > 2^(k+1) + (k+1)^2.

3^(k+1) = 3 ? 3^k
> 3(2^k + k^2) by the inductive hypothesis
= 3 ? 2^k + 3k^2
= 2 ? 2^k + 2^k + k^2 + 2k^2
= 2^(k+1) + 2^k + k^2 + 2k^2
? 2^(k+1) + 1 + k^2 + 2k
= 2^(k+1) + (k+1)^2

The second-to-last step makes use of the fact that for k ? 1, 2^k ? 1 and k^2 ? k.
Therefore 3^(k+1) > 2^(k+1) + (k+1)^2.

(b) For any n ? 1, the factorial function, denoted by n!, is the product of all the positive integers through n:
n! = 1 ? 2 ? 3 ? ? ? (n - 1) ? n

Prove that for n ? 4, n! ? 2^n.

Proving inequalities by induction.

Prove each of the following statements using mathematical induction.

(a) Prove that for n ? 2, 3^n > 2^n + n^2.

Solution
Proof.
By induction on n.

Base case: Show that for n = 2, 3^n > 2^n + n^2:
3^2 = 9 > 8 = 2^2 + 2^2, so the inequality holds for n = 2.

Inductive step: Assume that k is a positive integer and 3^k > 2^k + k^2. Show that 3^(k+1) > 2^(k+1) + (k+1)^2.

3^(k+1) = 3 ? 3^k
> 3(2^k + k^2) by the inductive hypothesis
= 3 ? 2^k + 3k^2
= 2 ? 2^k + 2^k + k^2 + 2k^2
= 2^(k+1) + 2^k + k^2 + 2k^2
? 2^(k+1) + 1 + k^2 + 2k
= 2^(k+1) + (k+1)^2

The second-to-last step makes use of the fact that for k ? 1, 2^k ? 1 and k^2 ? k.
Therefore 3^(k+1) > 2^(k+1) + (k+1)^2.

(b) For any n ? 1, the factorial function, denoted by n!, is the product of all the positive integers through n:
n! = 1 ? 2 ? 3 ? ? ? (n - 1) ? n

Prove that for n ? 4, n! ? 2^n.

Added by Michael M.

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