Math 104, Summer 2006, Homework 1 Solution
نویسنده
چکیده
Assignment: Section 1-1: 1.1, 1.4, 1.8, 1.12 Section 1-2: 2.3, 2.6 Section 1-3: 3.1, 3.2, 3.3, 3.6(b), 3.8 Section 1-1 1.1 Prove ∑n j=0 j 2 = 1 6n(n + 1)(2n + 1) for all natural numbers n. Proof. Let P(n) say ∑n j=0 j 2 = 6n(n + 1)(2n + 1). Then P(n) is a predicate in the variable n. (1) P(0) says ∑0 j=0 j 2 = 6 [0(0 + 1)(2 · 0 + 1)]. This is true since both sides evaluate to 0. (2) Let n ∈ N, and assume P(n) is true. Then ∑nj=0 j2 = 1 6n(n + 1)(2n + 1). We need to show P(n + 1) is true. P(n + 1) says ∑n+1 j=0 j 2 = 6 (n + 1)((n + 1) + 1)(2(n + 1) + 1). We have n+1 ∑
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