Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false:

1^2+ 4^2+ 7^2 +...+(3n-2)^2 = n(6n^2-3n-1) / 2

Answer 1

`\text{Proof by induction:}`

`\text{Test that the statement holds or n = 1}`


`LHS = (3 - 2)^(2) = 1`

`RHS = (6 - 4)/(2) = (2)/(2) = 1 = LHS`

`\text{Thus, the statement holds for the base case.}`


`\text{Assume the statement holds for some arbitrary term, n= k}`

`1^(2) + 4^(2) + 7^(2) + ... + (3k - 2)^(2) = (k(6k^(2) - 3k - 1))/(2)`


`\text{Prove it is true for n = k + 1}`

`RTP: 1^(2) + 4^(2) + 7^(2) + ... + [3(k + 1) - 2]^(2) = ((k + 1)[6(k + 1)^(2) - 3(k + 1) - 1])/(2) = ((k + 1)[6k^(2) + 9k + 2])/(2)`


`LHS = \underbrace{1^(2) + 4^(2) + 7^(2) + ... + (3k - 2)^(2)}_{(k(6k^(2) - 3k - 1))/(2)} + [3(k + 1) - 2]^(2)`

`= (k(6k^(2) - 3k - 1))/(2) + [3(k + 1) - 2]^(2)`

`= (k(6k^(2) - 3k - 1) + 2[3(k + 1) - 2]^(2))/(2)`

`= (k(6k^(2) - 3k - 1) + 2(3k + 1)^(2))/(2)`

`= (k(6k^(2) - 3k - 1) + 18k^(2) + 12k + 2)/(2)`

`= (k(6k^(2) - 3k - 1 + 18k + 12) + 2)/(2)`

`= \frac{k(6k^(2) + 15k + 11) + 2}{}`

`= ((k + 1)[6k^(2) + 9k + 2])/(2)`

`= ((k + 1)[6(k + 1)^(2) - 3(k + 1) - 1])/(2)`

`= RHS`

Since it is true for n = 1, n = k, and n = k + 1, by the principles of mathematical induction, it is true for all positive values of n.