Rewrite the right-hand side as a single fraction, and then factor the numerator completely.212()²2² +4² +6² + ... + (2(k+1))² =||2k(k + 1)(2k + 1)32(k + 11)(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()33) (2K2k +))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)3Let Pn denote the statement: 22 +4² +6² + ... + (2n)²+Check that P₁ is true: 2² = 4 and 2(CAssume Pk is true: 22 +42 +62 ++ (²2(2n)² = 2n(n + 1)(2n + 1)32² +4² +6² + ... + (2(k+1))²))( + ¹)(²1) (213To show that Pk+1 is true, add (2(k + 1))² to both sides of Pk.2² + 4² + 6² + ... + (2k)² + (2())²³=22|)²-²=+11)2k(k + 1)(2k + 1) 12(+3=Rewrite the right-hand side as a single fraction, and then factor the numerator completely.332k(k + 1)(2k + 1) +(2(3(k + 1)(2k + 1)Thus P₁ is true.22:))²

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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?

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter1: Fundamental Concepts Of Algebra

Section1.2: Exponents And Radicals

Problem 85E

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