Please help only with the box that has a red check mark next to it. Thank you. To show that Pk+1 is true, add (2(k+ 1))3 to both sides of Pk.+(2k)³ +(2k +123 +4³ +6³ ++6³ + ... + (2k)³ +Completely factor the right-hand side.23 + 43 + 6³ + ... + (2(k + 1))³ = 2( k +12k+12 k +1Rewrite the right-hand side in the desired form.3))³ = 2k²(k + 1)² + (2k + 1)²( 4 (k+1)) ² (k² + 4k +4)²(x + 223 + 4³ + 6³ + ... + (2(k + 1))³ = 2 (k+ 1)² ( (k+ 1) + 1)²X2+4k+1). ))³2)²

According to question,Given equation,23+ 43 + 63 + ....... + 2k3 + 2k+13 = 2k2k+12+ 2k+13

To show that Pk+1 is true, add (2(k+ 1))3 to both sides of Pk. +(2k)³ +(2k +1 23 +43 +63 + ... Completely factor the right-hand side. 23 +43 +63 +...+(2(k+ 1))³ = 2(k+1 = 2(k+1 = 2(k+1 Rewrite the right-hand side in the desired form. ))³ = 2k²(k + 1)² + (2【 k +1 1)²+(2(k+1,)) ],» )²(4(k+1) )²(k² + 4k+ 4 )²(x + 2 23 + 4³ + 6³ + ... + (2(k + 1))³ = 2 (k+ 1)² ( (k+1) + 1)² X + 4( k + 1 2+ ) 1,9 3² ))

To show that Pk+1 is true, add (2(k+ 1))3 to both sides of Pk. +(2k)³ +(2k +1 23 +43 +63 + ... Completely factor the right-hand side. 23 +43 +63 +...+(2(k+ 1))³ = 2(k+1 = 2(k+1 = 2(k+1 Rewrite the right-hand side in the desired form. ))³ = 2k²(k + 1)² + (2【 k +1 1)²+(2(k+1,)) ],» )²(4(k+1) )²(k² + 4k+ 4 )²(x + 2 23 + 4³ + 6³ + ... + (2(k + 1))³ = 2 (k+ 1)² ( (k+1) + 1)² X + 4( k + 1 2+ ) 1,9 3² ))

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...

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