Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false. I need help with the k+1 part of the proof

The base case is true, since

1² = 1

1 (6×1² - 3×1 - 1) / 2 = 1

Assume the equality holds for n = k, so that

1² + 4² + 7² + … + (3k - 2)² = k (6k ² - 3k - 1) / 2

Then for n = k + 1, we would have

1² + 4² + 7² + … + (3k - 2)² + (3 (k + 1) - 2)² = k (6k ² - 3k - 1) / 2 + (3 (k + 1) - 2)²

… = (6k ³ - 3k ² - k) / 2 + (3k + 3 - 2)²

… = (6k ³ - 3k ² - k) / 2 + (3k + 1)²

… = (6k ³ - 3k ² - k) / 2 + 9k ² + 6k + 1

… = (6k ³ - 3k ² - k + 18k ² + 12k + 2) / 2

… = (6k ³ + 15k ² + 11k + 2) / 2

We want to end up with

(k + 1) (6 (k + 1)² - 3 (k + 1) - 1) / 2

which means the numerator should have a factor of k + 1; it does, since

(6k ³ + 15k ² + 11k + 2) / (k + 1) = 6k ² + 9k + 2

This quotient should be equal to

6 (k + 1)² - 3 (k + 1) - 1

and it is; you can confirm this by expanding:

6 (k ² + 2k + 1) - 3 (k + 1) - 1 = 6k ² + 9k + 2

and we're done.

d

gimmee one of those gold crowns!

~coco

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