2 Consider the following partially correct proof that 1+2+...+n = n(n+1) for all n € J. Let P(n) be the proposition that

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2 Consider The Following Partially Correct Proof That 1 2 N N N 1 For All N J Let P N Be The Proposition That 1 (29.49 KiB) Viewed 31 times

2 Consider the following partially correct proof that 1+2+...+n = n(n+1) for all n € J. Let P(n) be the proposition that 1+2+...+ n = n(n+1) Basis Stop: Since 1 = 101), we see that P(1) is true. Inductive Step: Suppose that P(k) is true for some arbitrary k € J. Then 1+2+...+ k + (k+1) (k+ 1)(k+2) k(k+1) + (k + 1) = (k+ 1)(k+2) 2 2 k(k+1) 2(k+1) (k+ 1)(k+2) + 2 2 2 k(k+1)+2(k+1) (k+ 1)(k+2) 2 2 (k+ 1)(k+2) (k + 1)(k+2) 2 2 Thus, P(k+1) is true which proves that P(n) is true for all ne J by induction. Explain, using a few sentences, the error or errors that have been made in the above proof and why it or they are errors. Please be specific. You will be graded on how clearly and completely you make your case. on the identity.