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.

the simplified form is

step-by-step explanation:

we want to simplify the expression,

let us change the middle bar to a normal division sign to obtain,

we multiply the first fraction by the reciprocal of the second fraction to obtain,

we factor to obtain,

we cancel the common factors to get,

we simplify to get,

the correct answer is b

c

step-by-step explanation: