Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN: 9780470458365
Author: Erwin Kreyszig
Publisher: Wiley, John & Sons, Incorporated
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Use mathematical induction to prove the following statement.
Show that P
Proof (by mathematical induction): Let the property P(n) be the equation
1
(1₁ - 12/2) (1 - 3 2 ) ·- · (1₁ - 12/27)
=
We will show that P(n) is true for every integer n ≥
For every integer n ≥ 2,
1
2 ²₁ (1₁ - 2²2 ) (¹ - 2) · (1 - 227) -.
1
..
n²
Show that for each integer k ≥
1
1
(¹ - - 1 ) ( ¹ - - - ) ( ¹ - - - ) ... ( ¹.
1
1-
1
1 -
4²
and the right-hand side of P(k) is
We must show that P
is true: Before simplification, the left-hand side of P
1
(¹ - - - ) (¹² - - - ) ( ² - - ) ) -- - (¹1.
1-
1-
1
2²
3²
2k
[The inductive hypothesis is that the two sides of P(k) are equal.]
PC
Hence, P
n+ 1
2n
if P(k) is true, then P
2k
1
When the left- and right-hand sides of P
and the next-to-last factor in the left-hand side is 1-
1
1
1
1
1
1-
1
1-
(¹ - - ) (¹ - - ) (¹ - - ) ·- · (¹ - - ) (¹ - T
1-
1
+ 1
[+₁¹)
Before simplification, the right-hand side of P
1
is true. In other words, we must show that the left and right-hand sides of P
After substitution from the inductive hypothesis, the left-hand side of P
).(+-T
D²) - (
1
5)
1
n+ 1
2n
2k
2k(k+ 1)
is
7)
1
) 16 (₁-0²).
is 1
So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is
is true, which completes the inductive step.
2k
2 (K + 1)
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
+ 1
and the right-hand side is
becomes
1
are simplified, both can be shown to equal
+ 1
2.2
After simplification, both sides can be shown to equal
are equal. The left-hand side of P
is
Thus, P
is true.
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Transcribed Image Text:Use mathematical induction to prove the following statement. Show that P Proof (by mathematical induction): Let the property P(n) be the equation 1 (1₁ - 12/2) (1 - 3 2 ) ·- · (1₁ - 12/27) = We will show that P(n) is true for every integer n ≥ For every integer n ≥ 2, 1 2 ²₁ (1₁ - 2²2 ) (¹ - 2) · (1 - 227) -. 1 .. n² Show that for each integer k ≥ 1 1 (¹ - - 1 ) ( ¹ - - - ) ( ¹ - - - ) ... ( ¹. 1 1- 1 1 - 4² and the right-hand side of P(k) is We must show that P is true: Before simplification, the left-hand side of P 1 (¹ - - - ) (¹² - - - ) ( ² - - ) ) -- - (¹1. 1- 1- 1 2² 3² 2k [The inductive hypothesis is that the two sides of P(k) are equal.] PC Hence, P n+ 1 2n if P(k) is true, then P 2k 1 When the left- and right-hand sides of P and the next-to-last factor in the left-hand side is 1- 1 1 1 1 1 1- 1 1- (¹ - - ) (¹ - - ) (¹ - - ) ·- · (¹ - - ) (¹ - T 1- 1 + 1 [+₁¹) Before simplification, the right-hand side of P 1 is true. In other words, we must show that the left and right-hand sides of P After substitution from the inductive hypothesis, the left-hand side of P ).(+-T D²) - ( 1 5) 1 n+ 1 2n 2k 2k(k+ 1) is 7) 1 ) 16 (₁-0²). is 1 So, when the next-to-last factor is explicitly included in the expression for the left-hand side, the result is is true, which completes the inductive step. 2k 2 (K + 1) 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 + 1 and the right-hand side is becomes 1 are simplified, both can be shown to equal + 1 2.2 After simplification, both sides can be shown to equal are equal. The left-hand side of P is Thus, P is true.
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