Quiz Solutions for Math 6A, Spring Quarter 2003-04: Proofs and Sequences, Quizzes of Linear Algebra

Download Quiz Solutions for Math 6A, Spring Quarter 2003-04: Proofs and Sequences and more Quizzes Linear Algebra in PDF only on Docsity! Jim Lambers Math 6A Spring Quarter 2003-04 Quiz 6 Solution (Version A) 1. Prove by induction that n∑ i=1 i2 = n(n+ 1)(2n+ 1) 6 . Solution The basis step is n = 1, in which case the sum of the first 1 squares is equal to 1. We also have 1(1 + 1)(2(1) + 1) 6 = 1(2)(3) 6 = 1, so the formula holds in this case. For the inductive step, we assume that the formula holds for an arbitrary positive integer n, and show that it holds for n+ 1. That is, we must show that n+1∑ i=1 i2 = (n+ 1)((n+ 1) + 1)(2(n+ 1) + 1) 6 = (n+ 1)(n+ 2)(2n+ 3) 6 . Using our assumption that it holds for n, we have n+1∑ i=1 i2 = n∑ i=1 i2 + (n+ 1)2 = n(n+ 1)(2n+ 1) 6 + 6(n+ 1)2 6 = n(n+ 1)(2n+ 1) + 6(n+ 1)2 6 = (n+ 1)[n(2n+ 1) + 6(n+ 1)] 6 = (n+ 1)(2n2 + 7n+ 6) 6 = (n+ 1)(2n+ 3)(n+ 2) 6 . By the principle of mathematical induction, the formula holds for each positive integer n. 2. Find a formula for the terms of the sequence that begins with 2, 4, 7, 12, 21, 38, 71, 136, 265, 522, 1035. 1