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

Question

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.