Base case: the summation formula holds for n - 1 because both sides evaluate to 1/2. inductive step: suppose we already know for some arbitrary positive integer n that sigma ^n _k = 1 1/k)k + 1) = 1 - 1/n + 1. by adding the quantity 1/(n + 1)(n + 2)) to both sides, we get sigma ^n + 1 _k = 1 1/k(k + 1) = 1 - 1/n + 1 + 1//(n + 1)(n + 2) = 1 - n + 2 - 1//(n + 1)(n + 2) = 1 - n + 1//(n + 1)(n + 2) = 1 - 1/n + 2 thus, the summation formula holds for the case n + 1. this is the best of all the proofs in this quiz because it is not only correct, but also succinct and readable. it does not burden the reader with the unnecessary notational abstraction of a "p(n)". it's not ok in the base case to just state that both sides evaluate to 1/2. a proper informal proof must explain every last detail, even the details of basic arithmetic. an inductive proof is wrong if it does not refer to the statement to be proved as p(n). an inductive proof of a summation formula is not complete without a final declaration "thus, by the principle of mathematical induction, we have proved that the formula holds for all positive integers n." or words to that effect.