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base case: n = 1 1^3 + 2(1) = 3 3 | 3
assume that for k>=1, 3 | k^3+2k, we will prove that 3 | (k+1)^3+2(k+1) By inductive hypothesis, k^3+2k = 3m for some integer m, it follows that k^3 = 3m - 2k
(k+1)^3 + 2(k+1) = k^3+3k^2+5k+3 = (3m-2k) + 3k^2 + 5k + 3 by inductive hypothesis = 3m+3k^2+3k+3 = 3(m+k^2+k+1)
Since m and k are integers, m + k^2 + k + 1 is also an integer, therefore 3 | ((k+1)^3 + 2(k+1))
base case: b = 0(0+1) / 2 + 1 = 1
b = 1(1+1) / 2 + 1 = 2
b = 1 + 1 = 2
assume that for k >= 0, b = (k(k+1))/2 + 1, we will prove that b = ((k+1)(k+2))/2 + 1
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