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 Sk to be true, then
Sk: 6 + 12 + 18 + . . . + 6k = 3k(k + 1)
Step 3: Find Sk+1.
Sk+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 1 + 1)
Sk+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 2)
Step 4: Truth of Sk implies that Sk+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 Sk+1, it has been shown that the truth of Sk implies the truth of Sk+1 for every positive integer k. By the principle of mathematical induction, Sn is true for every positive integer n.
Mathematics
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