8.3 Applications of Series: Financial Models and Annuities

8.3 Applications of Series: Financial Models and Annuities

Quick Answer: A recurring savings plan (an ordinary annuity) compounds interest on each periodic deposit $d$ for a different number of periods, forming a finite geometric series. The total accumulated future value is computed using the NYSED Reference Sheet formula $S_n = \frac{a_1(1 - r^n)}{1 - r}$ with first term $a_1 = d$, common ratio $r = 1 + i$ (where periodic rate $i = \frac{r_{\text{annual}}}{m}$), and total periods $n = mt$. Similar geometric series model loan amortization, bouncing ball vertical travel, and repeated medical drug dosing.

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1. Financial Annuities as Geometric Series (AII-F.BF.7, AII-A.CED.1)

In standard compound interest, a single lump-sum principal $P$ is deposited once and left to accumulate interest over time according to $A = P\left(1 + \frac{r}{m}\right)^{mt}$. However, in authentic personal finance, individuals rarely invest a single lump sum; instead, they make recurring, equal contributions at regular intervals (e.g., monthly 401(k) retirement contributions, quarterly college savings deposits, or annual life insurance premiums).

A sequence of equal payments made at equal intervals of time is called an annuity. When payments are deposited at the end of each compounding period, it is termed an ordinary annuity.

Why Annuities Form a Geometric Series

Suppose an investor deposits an amount $d$ at the end of every compounding period into an account yielding an annual interest rate $r_{\text{annual}}$ compounded $m$ times per year. Let:

• $i = \frac{r_{\text{annual}}}{m}$ represent the periodic interest rate.
• $n = m \cdot t$ represent the total number of compounding periods over $t$ years.

Tracing the compound interest accrued on each individual deposit up to the end of the $n$-th period:

$$\begin{aligned} \text{Deposit } n \text{ (made at end of term):} & \quad d \\ \text{Deposit } n - 1 \text{ (compounds for 1 period):} & \quad d(1 + i)^1 \\ \text{Deposit } n - 2 \text{ (compounds for 2 periods):} & \quad d(1 + i)^2 \\ &\;\;\vdots \\ \text{Deposit } 1 \text{ (compounds for } n - 1 \text{ periods):} & \quad d(1 + i)^{n - 1} \end{aligned}$$

The total accumulated future value $FV$ is the sum of these individual compounded amounts, written in ascending order of powers:

$$FV = d + d(1 + i)^1 + d(1 + i)^2 + \dots + d(1 + i)^{n - 1}$$

This expression is precisely a finite geometric series where:

• Initial term: $a_1 = d$
• Common ratio: $r = 1 + i$
• Number of terms: $n$

Connecting to the NYSED Reference Sheet Formula

Applying the reference sheet formula $S_n = \frac{a_1(1 - r^n)}{1 - r}$ directly to this annuity series:

$$FV = \frac{d\left(1 - (1 + i)^n\right)}{1 - (1 + i)} = \frac{d\left(1 - (1 + i)^n\right)}{-i}$$

Multiplying the numerator and denominator by $-1$ transforms the equation into the standard financial formula for the future value of an ordinary annuity:

$$FV = \frac{d\left((1 + i)^n - 1\right)}{i}$$

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2. Loan Amortization and Mortgages

When a consumer borrows money to purchase a home (a mortgage) or automobile, the bank provides the loan principal $L$ upfront. The borrower repays the debt through $n$ equal monthly installments of size $M$.

From the lender's perspective, the initial loan amount $L$ must equal the present value of all future payments discounted by the monthly interest rate $i = \frac{r}{12}$:

$$L = \sum_{k=1}^{n} M(1 + i)^{-k} = M(1 + i)^{-1} + M(1 + i)^{-2} + \dots + M(1 + i)^{-n}$$

This is a finite geometric series where:

• First term: $a_1 = M(1 + i)^{-1} = \frac{M}{1 + i}$
• Common ratio: $r = (1 + i)^{-1} = \frac{1}{1 + i}$
• Number of terms: $n$

Summing this series yields the classical loan amortization formula:

$$L = M \left[ \frac{1 - (1 + i)^{-n}}{i} \right] \quad \Longleftrightarrow \quad M = L \left[ \frac{i}{1 - (1 + i)^{-n}} \right]$$

On Regents Part IV prompts, students are frequently provided this loan equation and asked to solve for the monthly payment $M$, calculate total interest paid ($n \cdot M - L$), or determine the loan balance remaining after a specified number of years.

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3. Physical and Biological Geometric Series Models

Geometric series model any dynamic system characterized by repeated fractional retention or decay.

The Bouncing Ball Model

When an elastic ball is dropped from an initial height $H$, it rebounds to a constant fraction $f$ of its previous height ($0 < f < 1$).

 Height
   |       Drop: H
   |         |   Bounce 1: 2·Hf
   |         |      /\    Bounce 2: 2·Hf²
   |         |     /  \       /\    Bounce 3: 2·Hf³
   |         v    /    \     /  \       /\
 --+---------+---+------+---+----+-----+--+----
           Ground

Calculating the total vertical distance traveled when the ball strikes the ground for the $n$-th time requires careful distinction between downward-only and two-way (up-and-down) motions:

1. The initial drop from height $H$ is strictly downward (distance $= H$).
2. Rebound 1 rises to $Hf$ and falls $Hf$ (distance $= 2Hf$).
3. Rebound 2 rises to $Hf^2$ and falls $Hf^2$ (distance $= 2Hf^2$).
4. Rebound $k$ rises and falls $Hf^k$ (distance $= 2Hf^k$).

When the ball hits the ground for the $n$-th time, it has completed the initial drop plus $(n - 1)$ full rebounds:

$$\text{Total Distance} = H + \sum_{k=1}^{n - 1} 2H f^k = H + 2Hf \left[ \frac{1 - f^{n - 1}}{1 - f} \right]$$

Pharmacokinetics: Repeated Drug Dosing Regimens

In pharmacology, patients receive regular doses of medication at uniform time intervals. Suppose a patient ingests a dose $D$ every 24 hours, and the body metabolizes and eliminates a fraction $(1 - c)$ of the drug daily, retaining a fraction $c$ ($0 < c < 1$).

Immediately after the $n$-th dose is administered:

• The newest dose contributes $D$.
• The previous dose has decayed to $D \cdot c$.
• The dose taken two days ago has decayed to $D \cdot c^2$.
• The first dose has decayed to $D \cdot c^{n - 1}$.

The total quantity of medication $A_n$ in the bloodstream immediately following the $n$-th dose is:

$$A_n = D + Dc + Dc^2 + \dots + Dc^{n-1} = \frac{D(1 - c^n)}{1 - c}$$

As the number of doses $n$ becomes very large, $c^n \to 0$, causing the medication level to level off at a stable plateau called the steady-state concentration:

$$\text{Steady-State Level} = \frac{D}{1 - c}$$

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4. Real-World Series Application Summary

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5. Worked Problem 1: Future Value of an Annuity (Regents Part IV Format)

Problem: At the end of each month, Marcus deposits $250 into a high-yield college savings account earning an annual interest rate of $4.8\%$ compounded monthly.

1. Write a geometric series expression representing the total account balance immediately after the 36th monthly deposit.
2. Algebraically determine the balance of the account at the end of 3 years (36 months), rounded to the nearest cent.
3. Determine the total amount of money Marcus contributed out-of-pocket, and calculate the total amount of compound interest earned over the 3-year period.

• Step 1: Calculate the periodic interest rate.

  $$i = \frac{r_{\text{annual}}}{m} = \frac{0.048}{12} = 0.004$$

• Step 2: Construct the geometric series. The 36th deposit earns 0 months of interest: $250$. The 35th deposit earns 1 month of interest: $250(1.004)^1$. The 1st deposit earns 35 months of interest: $250(1.004)^{35}$.

  $$S_{36} = 250 + 250(1.004)^1 + 250(1.004)^2 + \dots + 250(1.004)^{35} = \sum_{k=1}^{36} 250(1.004)^{k-1}$$

• Step 3: Evaluate using the geometric series sum formula. Using $a_1 = 250$, $r = 1.004$, and $n = 36$:

  $$S_{36} = \frac{250(1 - (1.004)^{36})}{1 - 1.004} = \frac{250(1 - 1.154552)}{-0.004} = \frac{250(-0.154552)}{-0.004}$$ $$S_{36} = \frac{-38.6381}{-0.004} \approx 9659.53$$

  The account balance after 3 years is $9,659.53.

• Step 4: Compute total deposits and interest earned.

  $$\text{Total Deposited} = 36 \times \$250 = \$9,000.00$$ $$\text{Interest Earned} = \text{Future Value} - \text{Total Deposited} = \$9,659.53 - \$9,000.00 = \$659.53$$

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6. Worked Problem 2: Bouncing Ball Total Distance (Regents Part III Format)

Problem: A tennis ball is dropped from a second-story window at a height of 28 feet above a concrete courtyard. On each bounce, the ball rebounds to $65\%$ of its previous height. Algebraically determine the total vertical distance the ball has traveled when it strikes the ground for the 6th time. Round your answer to the nearest tenth of a foot.

• Step 1: Decompose the motion.

  • Initial drop: $28$ ft (downward only).
  • The ball hits the ground for the 6th time after completing exactly $6 - 1 = 5$ rebounds.

• Step 2: Calculate the first rebound parameters. Height of first rebound: $H_1 = 28 \times 0.65 = 18.2$ ft. Distance of first rebound (up and down): $a_1 = 2 \times 18.2 = 36.4$ ft. Common ratio between successive rebounds: $r = 0.65$. Number of rebounds: $n = 5$.

• Step 3: Evaluate the sum of the 5 rebounds using the reference sheet formula.

  $$S_5 = \frac{a_1(1 - r^5)}{1 - r} = \frac{36.4(1 - (0.65)^5)}{1 - 0.65}$$ $$(0.65)^5 = 0.116029$$ $$S_5 = \frac{36.4(1 - 0.116029)}{0.35} = \frac{36.4(0.883971)}{0.35} = \frac{32.1765}{0.35} \approx 91.933 \text{ ft}$$

• Step 4: Add the initial drop height to obtain total vertical distance.

  $$\text{Total Distance} = 28 + 91.933 = 119.933 \approx 119.9 \text{ ft}$$

  The ball travels a total vertical distance of $119.9$ feet. Note that rounding to the nearest tenth keeps the digit 9 in the tenths place - writing $120.0$ is a rounding error, and rounding errors cost a credit under the NYSED rubric.

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7. Common Regents Pitfalls and Exam Strategies

• Pitfall 1: Confusing Annual Rate with Periodic Rate: Always divide the stated annual percentage rate by the compounding frequency: $i = \frac{r}{12}$ for monthly, $i = \frac{r}{4}$ for quarterly. Using $r = 0.06$ instead of $i = 0.005$ is a guaranteed deduction.
• Pitfall 2: Counting Rebounds on Bouncing Ball Prompts: If a prompt asks for the distance when the ball hits the ground for the $n$-th time, it has executed $(n - 1)$ rebounds. Setting the upper limit to $n$ adds an extra bounce that has not yet occurred.
• Pitfall 3: Premature Rounding in Financial Exponents: Do not round intermediate values like $(1.004)^{36} \approx 1.154552$ to $1.15$. Premature rounding compounds over large exponents and alters the final cent value, costing 1 credit under NYSED scoring keys.
• Part IV Justification Strategy: Always conclude extended constructed responses with a complete sentence that incorporates calculated numerical values, correct units (dollars, feet, hours), and clear mathematical reasoning.