Compute the values of the product (¹ + ²) (¹ + ²) (¹ + - ) ... (¹ + ²) 1+ 1+ for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture. (¹ + ³ ) (¹ + ² ) (¹ + ³ ) · · · (¹ + ²- ) [ 1+ Prove your conjecture by mathematical induction. Proof (by mathematical induction): Let P(n) be the equation (1+ ·( ¹ + ² ) ( ¹ + ² ) (¹ + — ) --- ( ¹ + ²+ ) [ Show that P(1) is true: Select P(1) from the choices below. O P(1) = 1 We will show that P(n) is true for every integer n 2 1. 0 (¹ + 1) (¹ + ²) (¹ + 3) (¹ + )-1+2+3 1+ 01+1-1+1 ○ P(1) -1 + The selected statement is true because both sides of the equation equal the same quantity. Show that for each integer kz 1, if P(k) is true, then P(K + 1) is true: Let k be any integer with k 2 1, and suppose that P(K) is true. The left-hand side of P(K) is (1 + 1) (¹ + 1) (¹ + 1) - ( [ [The inductive hypothesis is that the two sides of P(k) are equal.] We must show that P(x + 1) is true. The left-hand side of P(x + 1) is (1 + ) ( ¹ + ² ) ( ¹ + ² ) · · ( [ the left-hand and right-hand sides of P(k+ 1) are simplified, they both can be shown to equal and the right-hand side of P(k+ 1) is . Hence P(k+ 1) is true, which completes the inductive step. and the right-hand side of P(k) is . After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes (-Select-V 2). (1+2+1). Wh When

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Question

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Compute the values of the product
1
1
(¹ + ² ) (1 + ² ) (1 + ² ) ... (1 + ²)
1
n
for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture.
( ¹ + ² ) ( ¹ + ² ) ( ¹ + ² ) ... ( ¹ + ²¹) - [
1
1
1
1
=
Prove your conjecture by mathematical induction.
1
1
· (¹₁ + ² ) ( ¹₁ + ² ) ( ¹₁ + ² ) ... (¹₁ + ¹) = [
1
1
1
Proof (by mathematical induction): Let P(n) be the equation 1 +
Show that P(1) is true: Select P(1) from the choices below.
O P(1) = 1
1
1
1
O
0 ( ¹ + ² ) ( ¹₁ + ² ) ( ¹ + ² ) ( ¹ + ² ) = 1 + 2 + 3
1
1
1
1
01 + 1/ = 1 + 1
1
1
The selected statement is true because both sides of the equation equal the same quantity.
P(1) = 1 +
[The inductive hypothesis is that the two sides of P(k) are equal.]
Show that for each integer k ≥ 1, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + =
We must show that P(k+ 1) is true. The left-hand side of P(k + 1) is ( 1 +
) 15 ( 1₁ + ² +) ( ₁ + ²/2) ( ¹₁ + + + ) ). . . ( [
1
1
the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal
. We will show that P(n) is true for every integer n ≥ 1.
and the right-hand side of P(k + 1) is
5 ( ₁ + 1/ ) ( ₁ + / -) ( ₁ + / -) - - - (
1
[Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]
Hence P(k + 1) is true, which completes the inductive step.
and the right-hand side of P(k) is
After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes
1
1) · (₁ + x + ₁ )·\
1
k+ 1
---Select--- V
When
Transcribed Image Text:Compute the values of the product 1 1 (¹ + ² ) (1 + ² ) (1 + ² ) ... (1 + ²) 1 n for small values of n in order to conjecture a general formula for the product. Fill in the blank with your conjecture. ( ¹ + ² ) ( ¹ + ² ) ( ¹ + ² ) ... ( ¹ + ²¹) - [ 1 1 1 1 = Prove your conjecture by mathematical induction. 1 1 · (¹₁ + ² ) ( ¹₁ + ² ) ( ¹₁ + ² ) ... (¹₁ + ¹) = [ 1 1 1 Proof (by mathematical induction): Let P(n) be the equation 1 + Show that P(1) is true: Select P(1) from the choices below. O P(1) = 1 1 1 1 O 0 ( ¹ + ² ) ( ¹₁ + ² ) ( ¹ + ² ) ( ¹ + ² ) = 1 + 2 + 3 1 1 1 1 01 + 1/ = 1 + 1 1 1 The selected statement is true because both sides of the equation equal the same quantity. P(1) = 1 + [The inductive hypothesis is that the two sides of P(k) are equal.] Show that for each integer k ≥ 1, if P(k) is true, then P(k+ 1) is true: Let k be any integer with k ≥ 1, and suppose that P(k) is true. The left-hand side of P(k) is 1 + = We must show that P(k+ 1) is true. The left-hand side of P(k + 1) is ( 1 + ) 15 ( 1₁ + ² +) ( ₁ + ²/2) ( ¹₁ + + + ) ). . . ( [ 1 1 the left-hand and right-hand sides of P(k + 1) are simplified, they both can be shown to equal . We will show that P(n) is true for every integer n ≥ 1. and the right-hand side of P(k + 1) is 5 ( ₁ + 1/ ) ( ₁ + / -) ( ₁ + / -) - - - ( 1 [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.] Hence P(k + 1) is true, which completes the inductive step. and the right-hand side of P(k) is After substitution from the inductive hypothesis, the left-hand side of P(k+ 1) becomes 1 1) · (₁ + x + ₁ )·\ 1 k+ 1 ---Select--- V When
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