Answer
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 the same way.
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 $k=5$
