Question

Can someone check this mathematical induction problem?
Question and work below

Answer 1

1^`2 + 4^2 + 7^2 + ... + (3n - 2)^2 = \(\dfrac{n(6n^2-3n-1)}2\)

Answer 2

1^2 = \(\dfrac{1(6(1)^2 - 3(1)-1)}2\)

Answer 3

\(1^2 = \dfrac{1(36-3-1)}2\)

Answer 4

\(1^2 = \dfrac{36}2\)

Answer 5

\(1 \ne 16\)

Answer 6

mistake: trird post: 1*(6*(1)^2-3*(1)-1)=1*(6*1-6-1)

Answer 7

you got as 6*(1)^2=36 but that not true

Answer 8

oh whoops, thanks for catching that :)

Answer 9

np

Answer 10

So really it would be 2/2 = 1 which is true

Answer 11

yes that correct and true

Answer 12

Thank you ^_^

Answer 13

np