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10 questions from this quiz
Three
Base case
n(n+1)(2n+1)6\frac{n(n+1)(2n+1)}{6}6n(n+1)(2n+1)
111
(k+1)(k+2)(2k+3)6\frac{(k+1)(k+2)(2k+3)}{6}6(k+1)(k+2)(2k+3)
333
k3+2k=3mk^3 + 2k = 3mk3+2k=3m
k3+3k2+3k+1k^3 + 3k^2 + 3k + 1k3+3k2+3k+1
Not defining inductive hypothesis
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