Janet B. answered • 05/28/17

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Step 1: Show it's true for n = 1

1^3 + 2(1) = 3, which is divisible by 3

Step 2: Assume it's true for any integer k

Then k^3 + 2k = 3m, where m is an integer

Step 3: Show it's true for k + 1

(k+1)^3 + 2(k + 1) = k^3 + 3k^2 + 3k + 1 + 2k + 2

Substituting from Step 2, we have

3m + 3k^2 + 3k + 3

Factoring out 3, we have

3(m + k^2 + k + 1), which is a multiple of 3 and is therefore divisible by 3.