Example

Suppose we will deposit $100 each month into an account paying 6% interest. We assume that the account is compounded with the same frequency as we make deposits unless stated otherwise. Write an explicit formula that represents this scenario.

Answer: In this example:

• r = 0.06 (6%)
• k = 12 (12 compounds/deposits per year)
• d = $100 (our deposit per month)

Writing out the recursive equation gives [latex-display]{{P}_{m}}=\left(1+\frac{0.06}{12}\right){{P}_{m-1}}+100=\left(1.005\right){{P}_{m-1}}+100[/latex-display] Assuming we start with an empty account, we can begin using this relationship: [latex-display]P_0=0[/latex-display] [latex-display]P_1=(1.005)P_0+100=100[/latex-display] [latex-display]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[/latex-display] [latex-display]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[/latex-display] Continuing this pattern, after m deposits, we’d have saved: [latex-display]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[/latex-display] In other words, after m months, the first deposit will have earned compound interest for m-1 months. The second deposit will have earned interest for m­-2 months. The last month's deposit (L) would have earned only one month's worth of interest. The most recent deposit will have earned no interest yet. This equation leaves a lot to be desired, though – it doesn’t make calculating the ending balance any easier! To simplify things, multiply both sides of the equation by 1.005: [latex-display]1.005{{P}_{m}}=1.005\left(100{{\left(1.005\right)}^{m-1}}+100{{\left(1.005\right)}^{m-2}}+\cdots+100(1.005)+100\right)[/latex-display] Distributing on the right side of the equation gives [latex-display]1.005{{P}_{m}}=100{{\left(1.005\right)}^{m}}+100{{\left(1.005\right)}^{m-1}}+\cdots+100{{(1.005)}^{2}}+100(1.005)[/latex-display] Now we’ll line this up with like terms from our original equation, and subtract each side [latex-display]\begin{align}&\begin{matrix}1.005{{P}_{m}}&=&100{{\left(1.005\right)}^{m}}+&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&{}\\{{P}_{m}}&=&{}&100{{\left(1.005\right)}^{m-1}}+\cdots+&100(1.005)&+100\\\end{matrix}\\&\\\end{align}[/latex-display] Almost all the terms cancel on the right hand side when we subtract, leaving [latex-display]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex-display] Factor [latex]P_m[/latex] out of the terms on the left side. [latex-display]\begin{array}{c}P_m(1.005-1)=100{{\left(1.005\right)}^{m}}-100\\(0.005)P_m=100{{\left(1.005\right)}^{m}}-100\end{array}[/latex-display] Solve for P_m [latex-display]\begin{align}&0.005{{P}_{m}}=100\left({{\left(1.005\right)}^{m}}-1\right)\\&\\&{{P}_{m}}=\frac{100\left({{\left(1.005\right)}^{m}}-1\right)}{0.005}\\\end{align}[/latex-display] Replacing m months with 12N, where N is measured in years, gives [latex-display]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-1\right)}{0.005}[/latex-display] Recall 0.005 was r/k and 100 was the deposit d. 12 was k, the number of deposit each year.

Examples

A traditional individual retirement account (IRA) is a special type of retirement account in which the money you invest is exempt from income taxes until you withdraw it. If you deposit $100 each month into an IRA earning 6% interest, how much will you have in the account after 20 years?

Answer: In this example,

d = $100	the monthly deposit
r = 0.06	6% annual rate
k = 12	since we’re doing monthly deposits, we’ll compound monthly
N = 20	we want the amount after 20 years

Putting this into the equation: [latex-display]\begin{align}&{{P}_{20}}=\frac{100\left({{\left(1+\frac{0.06}{12}\right)}^{20(12)}}-1\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{20}}=\frac{100\left({{\left(1.005\right)}^{240}}-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(3.310-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(2.310\right)}{\left(0.005\right)}=\$46200 \\\end{align}[/latex-display] The account will grow to $46,200 after 20 years. Notice that you deposited into the account a total of $24,000 ($100 a month for 240 months). The difference between what you end up with and how much you put in is the interest earned. In this case it is $46,200 - $24,000 = $22,200.

This example is explained in detail here. https://youtu.be/quLg4bRpxPA  

Example

If you invest $100 each month into an account earning 3% compounded monthly, how long will it take the account to grow to $10,000?

Answer: This is a savings annuity problem since we are making regular deposits into the account.

d = $100	the monthly deposit
r = 0.03	3% annual rate
k = 12	since we’re doing monthly deposits, we’ll compound monthly

We don’t know N, but we want P_N to be $10,000. Putting this into the equation: [latex]10,000=\frac{100\left({{\left(1+\frac{0.03}{12}\right)}^{N(12)}}-1\right)}{\left(\frac{0.03}{12}\right)}[/latex]                        Simplifying the fractions a bit [latex-display]10,000=\frac{100\left({{\left(1.0025\right)}^{12N}}-1\right)}{0.0025}[/latex-display] We want to isolate the exponential term, 1.002512N, so multiply both sides by 0.0025 [latex]25=100\left({{\left(1.0025\right)}^{12N}}-1\right)[/latex]                             Divide both sides by 100 [latex]0.25={{\left(1.0025\right)}^{12N}}-1[/latex]                                     Add 1 to both sides [latex]1.25={{\left(1.0025\right)}^{12N}}[/latex]                                        Now take the log of both sides [latex]\log\left(1.25\right)=\log\left({{\left(1.0025\right)}^{12N}}\right)[/latex]                              Use the exponent property of logs [latex]\log\left(1.25\right)=12N\log\left(1.0025\right)[/latex]                          Divide by 12log(1.0025) [latex]\frac{\log\left(1.25\right)}{12\log\left(1.0025\right)}=N[/latex]                                               Approximating to a decimal N = 7.447 years It will take about 7.447 years to grow the account to $10,000.

This example is demonstrated here: https://youtu.be/F3QVyswCzRo

Example

After retiring, you want to be able to take $1000 every month for a total of 20 years from your retirement account. The account earns 6% interest. How much will you need in your account when you retire?

Answer: In this example,

d = $1000	the monthly withdrawal
r = 0.06	6% annual rate
k = 12	since we’re doing monthly withdrawals, we’ll compound monthly
N = 20	since were taking withdrawals for 20 years

We’re looking for P_0: how much money needs to be in the account at the beginning. Putting this into the equation: [latex-display]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-20(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-{{\left(1.005\right)}^{-240}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\times\left(1-0.302\right)}{\left(0.005\right)}=\$139,600 \\\end{align}[/latex-display] You will need to have $139,600 in your account when you retire. Notice that you withdrew a total of $240,000 ($1000 a month for 240 months). The difference between what you pulled out and what you started with is the interest earned. In this case it is $240,000 - $139,600 = $100,400 in interest.

View more about this problem in this video. https://youtu.be/HK2eRFH6-0U

Example

You can afford $200 per month as a car payment. If you can get an auto loan at 3% interest for 60 months (5 years), how expensive of a car can you afford? In other words, what amount loan can you pay off with $200 per month?

Answer: In this example,

d = $200	the monthly loan payment
r = 0.03	3% annual rate
k = 12	since we’re doing monthly payments, we’ll compound monthly
N = 5	since we’re making monthly payments for 5 years

We’re looking for P₀, the starting amount of the loan. [latex-display]\begin{align}&{{P}_{0}}=\frac{200\left(1-{{\left(1+\frac{0.03}{12}\right)}^{-5(12)}}\right)}{\left(\frac{0.03}{12}\right)}\\&{{P}_{0}}=\frac{200\left(1-{{\left(1.0025\right)}^{-60}}\right)}{\left(0.0025\right)}\\&{{P}_{0}}=\frac{200\left(1-0.861\right)}{\left(0.0025\right)}=\$11,120 \\\end{align}[/latex-display] You can afford a $11,120 loan. You will pay a total of $12,000 ($200 per month for 60 months) to the loan company. The difference between the amount you pay and the amount of the loan is the interest paid. In this case, you’re paying $12,000-$11,120 = $880 interest total.

Details of this example are examined in this video. https://youtu.be/5NiNcdYytvY  

Example

If a mortgage at a 6% interest rate has payments of $1,000 a month, how much will the loan balance be 10 years from the end the loan?

Answer: To determine this, we are looking for the amount of the loan that can be paid off by $1,000 a month payments in 10 years. In other words, we’re looking for P₀ when

d = $1,000	the monthly loan payment
r = 0.06	6% annual rate
k = 12	since we’re doing monthly payments, we’ll compound monthly
N = 10	since we’re making monthly payments for 10 more years

[latex-display]\begin{align}&{{P}_{0}}=\frac{1000\left(1-{{\left(1+\frac{0.06}{12}\right)}^{-10(12)}}\right)}{\left(\frac{0.06}{12}\right)}\\&{{P}_{0}}=\frac{1000\left(1-{{\left(1.005\right)}^{-120}}\right)}{\left(0.005\right)}\\&{{P}_{0}}=\frac{1000\left(1-0.5496\right)}{\left(0.005\right)}=\$90,073.45 \\\end{align}[/latex-display] The loan balance with 10 years remaining on the loan will be $90,073.45.

This example is explained in the following video: https://youtu.be/fXLzeyCfAwE

Example

A couple purchases a home with a $180,000 mortgage at 4% for 30 years with monthly payments. What will the remaining balance on their mortgage be after 5 years?

Answer: First we will calculate their monthly payments. We’re looking for d.

r = 0.04	4% annual rate
k = 12	since they’re paying monthly
N = 30	30 years
P0 = $180,000	the starting loan amount

We set up the equation and solve for d. [latex-display]\begin{align}&180,000=\frac{d\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-30(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&180,000=\frac{d\left(1-{{\left(1.00333\right)}^{-360}}\right)}{\left(0.00333\right)}\\&180,000=d(209.562)\\&d=\frac{180,000}{209.562}=\$858.93 \\\end{align}[/latex-display]   Now that we know the monthly payments, we can determine the remaining balance. We want the remaining balance after 5 years, when 25 years will be remaining on the loan, so we calculate the loan balance that will be paid off with the monthly payments over those 25 years.

d = $858.93	the monthly loan payment we calculated above
r = 0.04	4% annual rate
k = 12	since they’re doing monthly payments
N = 25	since they’d be making monthly payments for 25 more years

[latex-display]\begin{align}&{{P}_{0}}=\frac{858.93\left(1-{{\left(1+\frac{0.04}{12}\right)}^{-25(12)}}\right)}{\left(\frac{0.04}{12}\right)}\\&{{P}_{0}}=\frac{858.93\left(1-{{\left(1.00333\right)}^{-300}}\right)}{\left(0.00333\right)}\\&{{P}_{0}}=\frac{858.93\left(1-0.369\right)}{\left(0.00333\right)}=\$155,793.91 \\\end{align}[/latex-display] The loan balance after 5 years, with 25 years remaining on the loan, will be $155,793.91. Over that 5 years, the couple has paid off $180,000 - $155,793.91 = $24,206.09 of the loan balance. They have paid a total of $858.93 a month for 5 years (60 months), for a total of $51,535.80, so $51,535.80 - $24,206.09 = $27,329.71 of what they have paid so far has been interest.

More explanation of this example is available here: https://youtu.be/-J1Ak2LLyRo