Homework Help is Here – Start Your Trial Now!
Math
Calculus
Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 1)² 22 +42 +62+...+(2(k+1))² = = ... +(2(k+1))² = 2k(k+ 1)(2k + 1) 3 2(k+1) 2(k+ 1) (k+ Rewrite the right-hand side in the desired form. 22 +42 +62 +. 3 + 12 3 (2k(2k + 1) + 12 3 1) 3 )(2k + So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n. Need Help?

Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 1)² 22 +42 +62+...+(2(k+1))² = = ... +(2(k+1))² = 2k(k+ 1)(2k + 1) 3 2(k+1) 2(k+ 1) (k+ Rewrite the right-hand side in the desired form. 22 +42 +62 +. 3 + 12 3 (2k(2k + 1) + 12 3 1) 3 )(2k + So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n. Need Help?

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
See similar textbooks
icon
Related questions
Question
Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 2 12( )² 2² +4² +6² + ... + (2(k+1))² = || 2k(k + 1)(2k + 1) 3 2(k + 1 1)( 2(k + 1)(k+ Rewrite the right-hand side in the desired form. 2² +4² +6² + ... + (2(k+1))² = 3 + 3 (24(2k + 1) + 12( ) 3 3 ) (2K 2k + )) So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n.. Need Help?
Transcribed Image Text:Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 2 12( )² 2² +4² +6² + ... + (2(k+1))² = || 2k(k + 1)(2k + 1) 3 2(k + 1 1)( 2(k + 1)(k+ Rewrite the right-hand side in the desired form. 2² +4² +6² + ... + (2(k+1))² = 3 + 3 (24(2k + 1) + 12( ) 3 3 ) (2K 2k + )) So Pk+1 is true. We conclude by the principle of mathematical induction that Pn is true for all natural numbers n.. Need Help?
Use the principle of mathematical induction to show that the statement is true for all natural numbers. 2² +4² +6² + ... + (2n)² = 2n(n + 1)(2n + 1) 3 Let Pn denote the statement: 22 +4² +6² + ... + (2n)² + Check that P₁ is true: 2² = 4 and 2(C Assume Pk is true: 22 +42 +62 + + (² 2 (2n)² = 2n(n + 1)(2n + 1) 3 2² +4² +6² + ... + (2(k+1))² ))( + ¹)(² 1) (2 1 3 To show that Pk+1 is true, add (2(k + 1))² to both sides of Pk. 2² + 4² + 6² + ... + (2k)² + (2( ))²³ = 2 2 |)²-² = +1 1) 2k(k + 1)(2k + 1) 12( + 3 = Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 3 3 2k(k + 1)(2k + 1) +(2( 3 (k + 1)(2k + 1) Thus P₁ is true. 2 2 :))²
Transcribed Image Text:Use the principle of mathematical induction to show that the statement is true for all natural numbers. 2² +4² +6² + ... + (2n)² = 2n(n + 1)(2n + 1) 3 Let Pn denote the statement: 22 +4² +6² + ... + (2n)² + Check that P₁ is true: 2² = 4 and 2(C Assume Pk is true: 22 +42 +62 + + (² 2 (2n)² = 2n(n + 1)(2n + 1) 3 2² +4² +6² + ... + (2(k+1))² ))( + ¹)(² 1) (2 1 3 To show that Pk+1 is true, add (2(k + 1))² to both sides of Pk. 2² + 4² + 6² + ... + (2k)² + (2( ))²³ = 2 2 |)²-² = +1 1) 2k(k + 1)(2k + 1) 12( + 3 = Rewrite the right-hand side as a single fraction, and then factor the numerator completely. 3 3 2k(k + 1)(2k + 1) +(2( 3 (k + 1)(2k + 1) Thus P₁ is true. 2 2 :))²
Expert Solution
steps

Step by step

Solved in 3 steps with 3 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Precalculus
Precalculus
Calculus
ISBN:
9780135189405
Author:
Michael Sullivan
Publisher:
PEARSON
Calculus: Early Transcendental Functions
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning
SEE MORE TEXTBOOKS