
Mark M. answered 10/16/22
Tutor
5.0
(278)
Mathematics Teacher - NCLB Highly Qualified
13 + 2(1) = 3
Assume
k3 + 2k = 3n
Demonstrate
(k + 1)3 + 2(k + 1) = 3m
k3 + 3k2 + 3k + 1 + 2k + 2 = 3m
k3 + 2k + 3k2 + 3k + 3 = 3m
3n + 3(k2 + k + 1) = 3m
3(k2 + k + 1) = 3m - 3n
3(k2 + k + 1) = 3(m - n)
Q.E.D.