Using sequences with the Multiplier Effect example (Calculus 2)

Question

Here is the scenario: Anna, like all other citizens, gets a rebate of $300 and spends 90% of it. (this is called the marginal propensity to consume).

a. How do I find the pattern, using a table?

There is a table with 6 rows, 3 columns:

n column: for number of spending rounds

a_n column: for amounts spent each round --> a_0 is supposed to be $300.

S_n column: for cumulative sum of spending through round

(Sorry, I tried making a table, but it looked weird!)

It is hard for me to show what I've started without a functional table, but here goes:

Round One:

n = 0

a_n = $300

S_n = 300 ?

Round Two:

n = 1

a_n = 270 ? (90% of 300)

S_n = 270 + 300?

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b. Assuming the pattern above continues, what is the geometric series that models the total amount of money that moves through the economy beginning with the $300 payment?

Answer 1

Each round and following the "propensity to consume", 90% of the money we have is spent. The "money we have" comes from last round. So on round $i+1$ we spend 90% of the money that came from round $i $, that is, $a_{i+1} = 0.9a_i $. Applying that formula to itself, we get that $a_i = 0.9^ia_0 = 0.9^i*300$.

Now the cumulative sum of spending is given by summing over all the $a_i $. Therefore the total amount of money spent is given if we sum the $a_i $ for "infinite rounds":

total spent $= \sum_{i=0}^\infty a_i = \sum_{i=0}^\infty 0.9^ia_0$

which is a geometric series of ratio 0.9 and therefore can be summed up using the geometric series formula.

Assuming you are familiar with the formula itself, we have that

$$S_n = \sum_{i=0}^n 0.9^i*300 = 300\sum_{i=0}^n 0.9^i = 300\frac{1 - r^{n+1}}{1 - r}, r=0.9$$

Taking the limit $n \rightarrow \infty $

$$\lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} 300\frac{1 - r^{n+1}}{1 - r} = 300\frac {1}{1 - r}, r=0.9 $$

Since $0.9^i \rightarrow 0$ when $i \rightarrow \infty $.

Substituting we get

$$300\frac{1}{1 - \frac{9}{10}} = 300\frac{1}{\frac{1}{10}} = 300*10 = 3000$$.

Answer 2

I think you have found out that $a_n=300\cdot 0.9^n$

The n-th sum is

$S_n=\color{red}{300}+\color{blue}{300\cdot 0.9^1+300\cdot 0.9^2+300\cdot 0.9^3+\ldots+300\cdot 0.9^{n-1}+300\cdot 0.9^n}$

Multiplying both sides by $0.9$

$0.9\cdot S_n=\color{blue}{300\cdot 0.9^1+300\cdot 0.9^2+300\cdot 0.9^3+\ldots+300\cdot 0.9^{n-1}+300\cdot 0.9^n}+\color{red}{300\cdot 0.9^{n+1}}$

Now the second equation can be subtracted from the first equation. The terms in blue are equal in both equations. Only the red terms remains.

$S_n-0.9S_n=300-300\cdot 0.9^{n+1}$

Factoring out $S_n$ and $300$

$S_n(1-0.9)=300(1-0.9^{n+1})$

Dividing the equation by (1-0.9)

$S_n=300\cdot \frac{1-0.9^{n+1}}{1-0.9}=300\cdot \frac{1-0.9^{n+1}}{0.1}$

$n=0: \ \ S_0=300\cdot \frac{1-0.9^1}{0.1}=300$

$n=1:\ \ S_1=300\cdot \frac{1-0.9^2}{0.1}=300\cdot \frac{1-0.81}{0.9}=300\cdot \frac{0.19}{0.1}=570$

This matches your calculations.