Prove or disprove by math induction that 7+13+19+…+6n+1=n(3n+4) is true for ∀ n ∈ R

Question

step by step explanations would be helpful

Answer 1

Demonstrate true for n = 1

6(1) + 1 = 1(3(1) + 4

7 = 7

Assume true for n = n

7+13+19+...+ 6n + 1 = n(3n + 1)

Demonstrate true for n

7+13+19+...+ 6n + 1 = n(3n + 1)

Demonstrate true for n + 1

7+13+19+...+ (6n + 1) + [6(n + 1) + 1] = n(3n + 1) + [6(n + 1) + 1]

7+13+19+...+ (6n + 1) + [6(n + 1) + 1]= 3n² + n + 6n + 6 + 1

7+13+19+...+ (6n + 1) + [6(n + 1) + 1] = 3n² + 7n + 7

7+13+19+...+ (6n + 1) + [6(n + 1) + 1] = (n + 1)(3n + 4)

7+13+19+...+ (6n + 1) + [6(n + 1) + 1] = (n + 1)(3(n + 1) + 1)

QED

Answer 2

Prove the base case or the simplest case (n = 1)

When n = 1, we get 7 = 1*[3(1) + 4] which is true. So the base case is satisfied.

Now let's assume n = k is satisfied, and show that n = k+1 is also satisfied.

7 + 13 + 19 + ... + 6k+1 = k(3k+4).

Add in the k+1 term to both sides.

Therefore, 7 + 13 + 19 + ... + 6k+1 + 6(k+1) + 1 = k(3k+4) + 6(k+1) + 1.

Lastly, show that the right-hand side is what we would get for n = k+1.

k(3k+4) + 6(k+1) + 1 = 3k² + 4k + 6k + 6 + 1 = 3k² + 10k + 7 = (k+1)(3k + 7) = (k+1)[3(k+1) + 4]

This is what we'd expect the k+1 term to be, so therefore the k+1 term is satisfied.