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