Use mathematical induction to prove the following.6 + 12 + 18 + . . . + 6n = 3n(n + 1)

Question

What will be an ideal response?

Answer 1

Answers may vary. One possibility:
Sn: 6 + 12 + 18 + . . . + 6n = 3n(n + 1)
S1: 6 = 3 ? 1 ? (1 + 1)
Sk: 6 + 12 + 18 + . . . + 6k = 3k(k + 1)
Sk+1: 6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k + 1)(k + 2)
1. Basis step: Since 3 ? 1 ? (1 + 1) = 3 ? 2 = 6, S1 is true.
2. Induction step: Let k be any natural number. Assume Sk. Deduce Sk+1.
6 + 12 + 18 + . . . + 6k = 3k(k + 1) By Sk
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k(k + 1) + 6(k + 1) Adding 6(k + 1)
6 + 12 + 18 + . . . + 6k + 6(k + 1) = (3k + 6)(k + 1) Distributive law
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k + 2)(k + 1)
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k + 1)(k + 2).