Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 11 - Infinite Sequences and Series - 11.2 Series - 11.2 Exercises - Page 757: 73


a) $S_{n} = \frac{D(1-C^{n})}{1-C}$ b) $k=5$

Work Step by Step

a) Money spent by Government = D dollars. Money spent by people who receive D dollars. = a fraction C of D =CD Money spent by people who receive CD dollars. =a fraction C of CD =$C^{2}D$ Money spent by people who receive $C^{2}D$ dollars. =a fraction C of $C^{2}D$ =$C^{3}D$ Proceeding in the same way, we have Money spent by people in the $n$-th transaction. =a fraction C of $C^{n-2}D$ =$C^{n-1}D$ Therefore the total spending that has been generated after $n$ transactions. $S_{n} = D+CD+C^{2}D+C^{3}D+...C^{n-1}D$ $=D[1+C+C^{2}+C^{3}+...+C^{n-1}]$ $1+C+C^{2}+C^{3}+...+C^{n-1}$ is a geometric progression containing $n$ terms and with common ratio $C \lt 1$. Therefore, $1+C+C^{2}+C^{3}+...+C^{n-1} = \frac{1(1-C^{n})}{1-C}$ $=\frac{1-C^{n}}{1-C}$ Thus $S_{n} = \frac{D(1-C^{n})}{1-C}$ b) $S_{n} = \frac{D(1-C^{n})}{1-C}$ $\lim\limits_{n \to \infty}S_{n} = \lim\limits_{n \to \infty}\frac{D(1-C^{n})}{(1-C)}$ $=\frac{D(1-0)}{(1-C)}=\frac{D}{(1-C)}$ $=\frac{D}{S}$ [Since $C+S = 1 → S= 1-C$] $\lim\limits_{n \to \infty}S_{n} = kD$ Where, $k=\frac{1}{S}$ If the marginal propensity to consume is 80% i.e. $C=\frac{80}{100}=0.8$ $S=1-C=1-0.8=0.2=\frac{1}{5}$ And the multiplier $K=\frac{1}{S}=\frac{1}{\frac{1}{5}}=5$ Hence $k=5$
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