Prove the following statement by mathematical induction. For every integer n 2 0, 2" < (n + 2)!. Proof (by mathematical induction): Let P(n) be the inequality 2" < (n + 2)!. We will show that P(n) is true for every integer n 2 0. Show that P(o) is true: Before simplifying, the left-hand side of P(0) is and the right-hand side is The fact that the statement is true can be deduced from that fact that 2° = 1. Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. In other words, suppose that [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. P(k + 1) is the inequality . Information about P(k + 1) can be deduced from the following steps. Identify the reason for each step. 2k < (k + 2)! 2k. 2 < (k + 2)! · 2 2k + 1 < (k + 2)! · 2 2k + 1 < (k + 2)! · (k + 3) 2k + 1 < (k + 3)! by the induction hypothesis v by basic algebra ---Select--- ---Select--- ---Select--- Since 2k + 1 < (k + 3)! the inequality for P(k + 1) is --Select-- v , which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]

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Prove the following statement by mathematical induction.
For every integer n 2 0, 2" < (n + 2)!.
Proof (by mathematical induction): Let P(n) be the inequality 2" < (n + 2)!. We will show that P(n) is true for every integer n 2 0.
Show that P(0) is true: Before simplifying, the left-hand side of P(0) is
and the right-hand side is
The fact that the statement is true can
be deduced from that fact that 2° = 1.
Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. In other words, suppose that
[This is P(k), the inductive hypothesis.]
We must show that P(k + 1) is true. P(k + 1) is the inequality
. Information about P(k + 1) can be deduced from the following steps. Identify the reason for
each step.
2k < (k + 2)!
2k. 2 < (k + 2)! · 2
2k + 1 < (k + 2)! · 2
2k + 1 < (k + 2)! · (k + 3)
2k + 1
by the induction hypothesis v
by basic algebra
---Select---
---Select---
< (k + 3)!
---Select---
Since 2k + 1
< (k + 3)! the inequality for P(k + 1) is --Select--- v
which completes the inductive step.
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Transcribed Image Text:Prove the following statement by mathematical induction. For every integer n 2 0, 2" < (n + 2)!. Proof (by mathematical induction): Let P(n) be the inequality 2" < (n + 2)!. We will show that P(n) is true for every integer n 2 0. Show that P(0) is true: Before simplifying, the left-hand side of P(0) is and the right-hand side is The fact that the statement is true can be deduced from that fact that 2° = 1. Show that for each integer k 2 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k 2 0, and suppose that P(k) is true. In other words, suppose that [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. P(k + 1) is the inequality . Information about P(k + 1) can be deduced from the following steps. Identify the reason for each step. 2k < (k + 2)! 2k. 2 < (k + 2)! · 2 2k + 1 < (k + 2)! · 2 2k + 1 < (k + 2)! · (k + 3) 2k + 1 by the induction hypothesis v by basic algebra ---Select--- ---Select--- < (k + 3)! ---Select--- Since 2k + 1 < (k + 3)! the inequality for P(k + 1) is --Select--- v which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
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