Ch9 Sequences, Series, And Probability Exam Prep - Test Bank | College Algebra 5e by Young by Cynthia Y. Young. DOCX document preview.

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Ch9 Sequences, Series, And Probability Exam Prep

College Algebra, 5e (Young)

Chapter 9 Sequences, Series, and Probability

9.4 Mathematical Induction

1) Prove the statement using mathematical induction for all positive integers, n.

(n) with superscript (3) ≤ (n) with superscript (4)

(1) with superscript (3) ≤ (1) with superscript (4)

If we assume this is true for n = k, then for k + 1:

(k) with superscript (3) + 3(k) with superscript (3) + ... + 1 is compared to (k) with superscript (4) + 6(k) with superscript (3) + ... + 1.

Since (k) with superscript (3) ≤ (k) with superscript (4) and (k) with superscript (2) ≤ (k) with superscript (3) , .... we can conclude that ((k + 1)) with superscript (3) ≤ ((k + 1)) with superscript (4).

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

2) Prove the statement using mathematical induction for all positive integers, n.

(7) with superscript (n) < (7) with superscript (n+1)

(7) with superscript (1) < (7) with superscript (2)

Assuming that (7) with superscript (k) < (7) with superscript (k+1)

Then,

(7) with superscript (k+1) = 7 ∙ (7) with superscript (1)

(7) with superscript (k+2) = 7 ∙ (7) with superscript (2)

Since 7 > 1, then

7 ∙ (7) with superscript (1) < 7 ∙ (7) with superscript (2)

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

3) Prove the statement using mathematical induction for all positive integers, n.

8 + 13 + 18 + ... + (5n + 3) = n(5n + 11)/2

5n + 3 = 5 · 1 + 3 = 8

n(5n + 11)/2 = 1(5 ∙ 1 + 11)/2 = 8

Assuming it is true of n = k, then for k + 1,

8 + 13 + 18 + ... + (5k + 3) + (5(k + 1) + 3) = (k + 1)(5(k + 1) + 11)/2

Since

8 + 13 + 18 + ... + (5k + 3) = k(5k + 11)/2

and:

8 + 13 + 18 + ... + (5k + 3) + (5(k + 1) + 3) = k (5k + 11)/2 + (5(k + 1) + 3)

k(5k + 11)/2 + (5(k + 1) + 3) = (5(k) with superscript (2) + 11k)/2 + 5k + 5+ 3

(5(k) with superscript (2) + 11k)/2 + 5k + 5+ 3 = (k + 1)(5(k + 1)+ 11)/2

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

4) Prove the statement using mathematical induction for all positive integers, n.

(9) with superscript (0) + (9) with superscript (1) + (9) with superscript (2) + ... + (9) with superscript (n-1) = ((9) with superscript (n) - 1)/8

(9) with superscript (0) = 1

and

((9) with superscript (n) - 1)/8 = 1

If we assume it is true for n = k, then, for n = k + 1:

(9) with superscript (0) + (9) with superscript (1) + (9) with superscript (2) + ... + (9) with superscript (k-1) + (9) with superscript (k) = ((9) with superscript (k+1) - 1)/8

and:

((9) with superscript (k) - 1)/8 + (9) with superscript (k) = ((9) with superscript (k+1) - 1)/8

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

5) Prove the statement using mathematical induction for all positive integers, n.

(1/2) + (1/4) + (1/64) + ... + (1/(2) with superscript (n)) = 1 - (1/(2) with superscript (n))

(1/2) = 1- (1/2) = (1/2)

Assuming it is true for n = k, then, for n = k + 1:

(1/2) + (1/4) + (1/64) + ... + (1/(2) with superscript (k)) + (1/(2) with superscript (k+1)) = 1 - (1/(2) with superscript (k+1))

and:

1 - (1/(2) with superscript (k)) + (1/(2) with superscript (k+1)) = 1 - (1/(2) with superscript (k+1))

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

6) Prove the statement using mathematical induction for all positive integers, n.

(1/5) + (1/10) + (1/1,000) + ... + (1/(5) with superscript (n)) = (1/4)(1 - (1/(5) with superscript (n)))

(1/5) = (1/4)(1 - (1/(5) with superscript (1))) = (1/5)

If you assume it is true for n = k, then, for n = k + 1:

(1/5) + (1/10) + (1/1,000) + ... + (1/(5) with superscript (k)) + (1/(5) with superscript (k+1)) = (1/4)(1 - (1/(5) with superscript (k+1)))

and:

(1/4)(1 - (1/(5) with superscript (k))) + (1/(5) with superscript (k+1)) = (1/4)(1 - (1/(5) with superscript (k+1)))

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

7) Prove the statement using mathematical induction for all positive integers, n.

n(n + 1)(n + 2) is divisible by 3

1(1 + 1)(1 + 2) = 6 which is divisible by 3

If you assume it is true for n = k; then, for n = k + 1:

(k + 1)((k + 1) + 1)((k + 1) + 2)

If

k(k + 1)(k + 2) is divisible by 3 then (k + 1)((k + 1) + 1)((k + 1) + 2) is divisible by 3

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

8) Prove the statement using mathematical induction for all positive integers, n.

(1 ● 2) + (2 ● 3) + (3 ● 4) + ... + n(n + 1) = (n(n + 1)( n + 2)/3)

1(1 + 1) = (1(1 + 1)( 1 + 2)/3)

Since:

1(1 + 1) = 2 and (1(1 + 1)( 1 + 2)/3) = 2

If we assume it is true for n = k, then, for n = k + 1:

(1 ● 2) + (2 ● 3) + (3 ● 4) + ... + k(k + 1) + (k + 1)((k + 1) + 1) =

((k + 1)((k + 1)( (k + 1) + 2)/3)

and

(k(k + 1)( (k + 2)/3) + (k + 1)((k +1) + 1) = ((k + 1)((k + 1) + 1)( (k + 1) + 2)/3)

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

9) Prove the statement using mathematical induction for all positive integers, n.

(1/2 ● 3) + (1/3 ● 4) (1/4 ● 5) + ... + (1/(n + 1)(n + 2)) = (n/2(n + 2))

(1/(1 + 1)(1 +2)) = (1/2(1 + 2)) = (1/6)

If we assume it is true for n = k, then, for n = k + 1:

(1/2 ● 3) + (1/3 ● 4) (1/4 ● 5) + ... + (1/(k + 1)(k + 2)) + (1/((k + 1) + 1)((k + 1) + 2) + (k + 1/2((k + 1) + 2))

and

(1/2(k + 2)) + (1/((k + 1) + 1)((k + 1) + 2) = (k + 1/2((k + 1) + 2))

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

10) Prove the statement using mathematical induction for all positive integers, n.

(1) with superscript ( ) + (2) with superscript ( ) + (3) with superscript ( ) + ... + (n) with superscript ( ) = ((n) with superscript ( )((n + 1)) with superscript ( ) /2)

(1) with superscript ( ) = ((1) with superscript ( )((1 + 1)) with superscript ( ) /2) = 1

If we assume it is true for n = k, then, for n = k + 1:

(1) with superscript ( ) + (2) with superscript ( ) + (3) with superscript ( ) + ... + (k) with superscript ( ) + ((k + 1)) with superscript ( ) = (((k + 1)) with superscript ( )(((k + 1) + 1)) with superscript ( ) /2)

and:

((k ) with superscript ( )((k + 1)) with superscript ( ) /2) + ((k + 1)) with superscript ( ) = (((k + 1)) with superscript ( )(((k + 1) + 1)) with superscript ( ) /2)

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

11) Which relationship below can be shown to be true through mathematical induction?

A) (k) with superscript (n) ≤ (k) with superscript (n+2), for n = 1, 3, 5,..., 2n - 1 and k = all real numbers.

B) (k) with superscript (n) ≤ (k) with superscript (n+2), for n ≥ 1 and k = all real numbers.

C) (k) with superscript (n) ≤ (k) with superscript (n+2), for n = 2, 4, 6,..., 2l and k = all real numbers.

D) (k) with superscript (n) ≤ (k) with superscript (n+2), for n = 1, 3, 5,..., 2n - 1 and k < -1.

Diff: 2 Var: 1

Chapter/Section: Ch 09, Sec 04

Learning Objective: Prove mathematical statements using mathematical induction.

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Document Type:
DOCX
Chapter Number:
9
Created Date:
Aug 21, 2025
Chapter Name:
Chapter 9 Sequences, Series, And Probability
Author:
Cynthia Y. Young

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