Stewart - Calculus - Early Transcendentals 6e HQ (Thomson, 2008) Episode 13

Tham khảo tài liệu 'stewart - calculus - early transcendentals 6e hq (thomson, 2008) episode 13', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | APPENDIX E SIGMA NOTATION A35 2 THEOREM If c is any constant that is it does not depend on i then n n a 2 cai c s di i m i m n n n c 2 a - bi 2 ai - 2 bi i m i m i m n n n b 2 a bò 2 ai 2 b i m i m i m PROOF To see why these rules are true all we have to do is write both sides in expanded form. Rule a is just the distributive property of real numbers cam cam 1 can c am am 1 an Rule b follows from the associative and commutative properties flm bm flm 1 bm 1 fln bn flm am 1 an bm bm 1 bn Rule c is proved similarly. EXAMPLE 3 Find 1. i 1 SOLUTION 1 1 1 1 n i 1 v V n terms EXAMPLE 4 Prove the formula for the sum of the first n positive integers . n n 1 2 i 1 2 3 n ----------------- i-1 2 SOLUTION This formula can be proved by mathematical induction see page 77 or by the following method used by the German mathematician Karl Friedrich Gauss 1777-1855 when he was ten years old. Write the sum 5 twice once in the usual order and once in reverse order 1 2 3 n 1 n s n n 1 n 2 2 1 Adding all columns vertically we get 2S n 1 n 1 n 1 n 1 n 1 On the right side there are n terms each of which is n 1 so n n 1 2S n n 1 or s 2 EXAMPLE 5 Prove the formula for the sum of the squares of the first n positive integers n n 1 2n 1 2 i 12 22 32 n2 i-1 6 A36 APPENDIX E SIGMA NOTATION SOLUTION 1 Let S be the desired sum. We start with the telescoping sum or collapsing sum Most terms cancel in pairs. n s 1 i 3 - i3 23 - 13 33 - 23 43 - 33 . n 1 3 - n3 i 1 n 1 3 - 13 n 3 3n2 3n On the other hand using Theorem 2 and Examples 3 and 4 we have n n n n n 2 1 i 3 i 3 s 3i2 3i 1 3 s i2 3 s i s 1 i 1 i 1 i 1 i 1 i 1 3S 3 n n 1 n 3S 2 n2 2n Thus we have n3 3n2 3n 3S 2 n2 2 n Solving this equation for S we obtain 3S n3 I n2 1 n 2n3 3n2 n n n 1 2n 1 or S ---------A ------- -------A---------- 6 6 PRINCIPLE OF MATHEMATICAL INDUCTION Let Sn be a statement involving the positive integer n. Suppose that 1. S1 is true. 2. If Sk is true then Sk 1 is true. Then Sn is true for all positive integers n. SOLUTION 2 .