Use the principle of mathematical induction to show that the mathematical statement is true for all natural numbers n.Sn: 6 + 12 + 18 + . . . + 6n = 3n(n + 1)
What will be an ideal response?
| S1: | 6  (3 ? 1)(1 + 1) |
3 ? 26 = 6 ?
Sk: 6 + 12 + 18 + . . . + 6k = 3k(k + 1)
Sk+1: 6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 2)
We work with Sk. Because we assume that Sk is true, we add the next multiple of 6, namely
6(k+1), to both sides.
6 + 12 + 18 + . . . + 6k + 6(k + 1) = 3k(k + 1) + 6(k + 1)
6 + 12 + 18 + . . . + 6(k + 1) = (k + 1)(3k + 6)
6 + 12 + 18 + . . . + 6(k + 1) = 3(k + 1)(k + 2)
We have shown that if we assume that Sk is true, and we add 6(k+1) to both sides of Sk, then Sk+1 is also true. By the principle of mathematical induction, the statement Sn is true for every positive integer n.
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