Discrete Mathematics and Its Applications, Seventh Edition

Published by McGraw-Hill Education
ISBN 10: 0073383090
ISBN 13: 978-0-07338-309-5

Chapter 2 - Section 2.4 - Sequences and Summations - Exercises - Page 168: 20

Answer

a. Pn-1 (1,011) b. 1,011^n x P0 c. P20 = 1,011^20 x 6,9B

Work Step by Step

a. r = \frac{11}{1000} P0 = 6,9 B P1 = 6,9 B (1+\frac{11}{1000}) P2 = P1 (1+\frac{11}{1000}) = 6,9 B (1+\frac{11}{1000}) (1+\frac{11}{1000}) = 6,9 (1+\frac{11}{1000})^{2} Pn = Pn-1 (1+\frac{11}{1000}) Pn = Pn-1 x 1,011 b. Pn = Pn-1 x 1,011 P0 = 6,9 B P1 = 1,011 x P0 = 1,011 x 6,9 B P2 = 1,011 x P1 = 1,011 x (1,011 x 6,9 B) = 1,011^2 x 6,9 B Pn = 1,011^{n} x 6,9 B Pn = 1,011^{n} x P0 c. 2010 ----> n = 0 2030 ----> n = 2030 - 2010 2030 ----> n = 20 Pn = 1,011^{n} x P0 Pn = 1,011^{20} x 6,9 B
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