A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely. Show your work.

Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

Sk: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + k(k + 1) = [k(k + 1)(k + 2)]/3

Sk+1: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + (k+1)(k+1 + 1) = [(k+1)(k+1 + 1)(k+1 + 2)]/3

Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k+1)(k+1 + 1)

= [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k+1)((k+1) + 1)

= [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + k((k+1) + 1)+1((k+1) + 1)

= [(k+1)(k+1 + 1)(k+1 + 2)]/3 Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + k(k+1) + k + 1(k+1) + 1 = [(k+1)(k+1 + 1)(k+1 + 2)]/3

So this proves that Sn: 1 · 2 + 2 · 3 + 3 · 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3 for all natural numbers n.

To simplify Sk+1 completely, we can distribute the terms and then combine like terms:

Sk+1: 1 · 2 + 2 · 3 + 3 · 4 + . . . + (k+1)(k+1 + 1)

= (1 · 2 + 2 · 3 + 3 · 4 + . . . + k(k+1)) + (k + 1)(k + 1 + 1)

= Sk + (k + 1)(k + 2)

Now we can substitute Sk back into the equation:

Sk+1 = [k(k + 1)(k + 2)]/3 + (k + 1)(k + 2)

= [(k(k + 1)(k + 2) + 3(k + 1)(k + 2))/3]

= [(k + 1)(k(k + 2) + 3(k + 2))/3]

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

= [(k + 1)(k^2 + 5k + 6)/3]

Therefore, we have simplified Sk+1 completely.