Use mathematical induction to prove the following statement. For every integer n ≥ 2, (1-2) (1-3)... (1-2)= n + 1 Proof (by mathematical induction): Let the property P(n) be the equation (1 - 237) (1-33) ··· (1 - 1327) We will show that P(n) is true for every integer n ≥ n+1 2n 2n +1 Show that P is true: Before simplification, the left-hand side of is 1- and the right-hand side is After simplification, both sides can be shown to equal 2.2 Show that for each integer k ≥ if P(k) is true, then P 1 (1-2) (1-3) (1-2) ... (1- and the right-hand side of P(k) is 2k [The inductive hypothesis is that the two sides of P(k) are equal.] We must show that Pl (1-2) (1-3) (1-2) ... is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is is true. In other words, we must show that the left- and right-hand sides of P are equal. The left-hand side of P and the next-to-last factor in the left-hand side is 1 (1-12) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is 1 1- 1 (-) (-)(-)-(-)(( After substitution from the inductive hypothesis, the left-hand side of P 2k Before simplification, the right-hand side of P When the left- and right-hand sides of P Hence, P 2k 2k 2k(k+1) IS +1 1 becomes 2(k + 1) are simplified, both can be shown to equal is true, which completes the inductive step. Thus, P is true.
Use mathematical induction to prove the following statement. For every integer n ≥ 2, (1-2) (1-3)... (1-2)= n + 1 Proof (by mathematical induction): Let the property P(n) be the equation (1 - 237) (1-33) ··· (1 - 1327) We will show that P(n) is true for every integer n ≥ n+1 2n 2n +1 Show that P is true: Before simplification, the left-hand side of is 1- and the right-hand side is After simplification, both sides can be shown to equal 2.2 Show that for each integer k ≥ if P(k) is true, then P 1 (1-2) (1-3) (1-2) ... (1- and the right-hand side of P(k) is 2k [The inductive hypothesis is that the two sides of P(k) are equal.] We must show that Pl (1-2) (1-3) (1-2) ... is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is is true. In other words, we must show that the left- and right-hand sides of P are equal. The left-hand side of P and the next-to-last factor in the left-hand side is 1 (1-12) So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is 1 1- 1 (-) (-)(-)-(-)(( After substitution from the inductive hypothesis, the left-hand side of P 2k Before simplification, the right-hand side of P When the left- and right-hand sides of P Hence, P 2k 2k 2k(k+1) IS +1 1 becomes 2(k + 1) are simplified, both can be shown to equal is true, which completes the inductive step. Thus, P is true.
College Algebra (MindTap Course List)
12th Edition
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:R. David Gustafson, Jeff Hughes
Chapter8: Sequences, Series, And Probability
Section8.5: Mathematical Induction
Problem 42E
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![Use mathematical induction to prove the following statement.
For every integer n ≥ 2,
(1-2) (1-3)... (1-2)=
n + 1
Proof (by mathematical induction): Let the property P(n) be the equation
(1 - 237) (1-33) ··· (1 - 1327)
We will show that P(n) is true for every integer n ≥
n+1
2n
2n
+1
Show that P
is true: Before simplification, the left-hand side of
is
1-
and the right-hand side is
After simplification, both sides can be shown to equal
2.2
Show that for each integer k ≥
if P(k) is true, then P
1
(1-2) (1-3) (1-2)
...
(1-
and the right-hand side of P(k) is
2k
[The inductive hypothesis is that the two sides of P(k) are equal.]
We must show that Pl
(1-2) (1-3) (1-2)
...
is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is
is true. In other words, we must show that the left- and right-hand sides of P
are equal. The left-hand side of P
and the next-to-last factor in the left-hand side is
1
(1-12)
So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is
1
1-
1
(-) (-)(-)-(-)((
After substitution from the inductive hypothesis, the left-hand side of P
2k
Before simplification, the right-hand side of P
When the left- and right-hand sides of P
Hence, P
2k
2k
2k(k+1)
IS
+1
1
becomes
2(k + 1)
are simplified, both can be shown to equal
is true, which completes the inductive step.
Thus, P
is true.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3cf874b-7a7b-478f-a7b8-421442e72224%2F0de07d99-e2e2-4acb-a869-acdd1703dd89%2F4s5o9h8_processed.png&w=3840&q=75)
Transcribed Image Text:Use mathematical induction to prove the following statement.
For every integer n ≥ 2,
(1-2) (1-3)... (1-2)=
n + 1
Proof (by mathematical induction): Let the property P(n) be the equation
(1 - 237) (1-33) ··· (1 - 1327)
We will show that P(n) is true for every integer n ≥
n+1
2n
2n
+1
Show that P
is true: Before simplification, the left-hand side of
is
1-
and the right-hand side is
After simplification, both sides can be shown to equal
2.2
Show that for each integer k ≥
if P(k) is true, then P
1
(1-2) (1-3) (1-2)
...
(1-
and the right-hand side of P(k) is
2k
[The inductive hypothesis is that the two sides of P(k) are equal.]
We must show that Pl
(1-2) (1-3) (1-2)
...
is true: Let k be any integer with k ≥ 2, and suppose that P(k) is true. Before any simplification, the left-hand side of P(k) is
is true. In other words, we must show that the left- and right-hand sides of P
are equal. The left-hand side of P
and the next-to-last factor in the left-hand side is
1
(1-12)
So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is
1
1-
1
(-) (-)(-)-(-)((
After substitution from the inductive hypothesis, the left-hand side of P
2k
Before simplification, the right-hand side of P
When the left- and right-hand sides of P
Hence, P
2k
2k
2k(k+1)
IS
+1
1
becomes
2(k + 1)
are simplified, both can be shown to equal
is true, which completes the inductive step.
Thus, P
is true.
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