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 correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2