Use mathematical induction to prove the statement is true for all positive integers n.6 + 12 + 18 + . . . + 6n = 3n(n + 1)

What will be an ideal response?

Step 1:

Determine if S₁, S₂, and S₃ are true.

Step 2: Assume S_k to be true, then
   S_k: 6 + 12 + 18 + . . . + 6k = 3k(k + 1)

Step 3: Find S_k+1.
   S_k+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 1 + 1)
   S_k+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 2)

Step 4: Truth of S_k implies that S_k+1 is true.
   6 + 12 + 18 + . . . + 6k = 3k(k + 1)
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k(k + 1) + 6(k + 1)
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k² + 3k + 6k + 6
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k² + 9k + 6
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k² + 3k + 2)
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3(k + 1)(k + 2)

Since the last equation is S_k+1, it has been shown that the truth of S_k implies the truth of S_k+1 for every positive integer k. By the principle of mathematical induction, S_n is true for every positive integer n.

Mathematics