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]{{P}_{m}}=\left(1+\frac{0.06}{12}\right){{P}_{m-1}}+100=\left(1.005\right){{P}_{m-1}}+100[/latex]

Assuming we start with an empty account, we can begin using this relationship:

[latex]P_0=0[/latex]

[latex]P_1=(1.005)P_0+100=100[/latex]

[latex]P_2=(1.005)P_1+100=(1.005)(100)+100=100(1.005)+100[/latex]

[latex]P_3=(1.005)P_2+100=(1.005)(100(1.005)+100)+100=100(1.005)^2+100(1.005)+100[/latex]

Continuing this pattern, after m deposits, we’d have saved:

[latex]P_m=100(1.005)^{m-1}+100(1.005)^{m-2} +L+100(1.005)+100[/latex]

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]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]

Distributing on the right side of the equation gives

[latex]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]

Now we’ll line this up with like terms from our original equation, and subtract each side

[latex]\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]

Almost all the terms cancel on the right hand side when we subtract, leaving

[latex]1.005{{P}_{m}}-{{P}_{m}}=100{{\left(1.005\right)}^{m}}-100[/latex]

Factor [latex]P_m[/latex] out of the terms on the left side.

[latex]\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]

Solve for P_m

[latex]\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]

Replacing m months with 12N, where N is measured in years, gives

[latex]{{P}_{N}}=\frac{100\left({{\left(1.005\right)}^{12N}}-1\right)}{0.005}[/latex]

Recall 0.005 was r/k and 100 was the deposit d. 12 was k, the number of deposit each year.

In this example,

Putting this into the equation:

[latex]\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}}\approx\frac{100\left(3.310204476-1\right)}{\left(0.005\right)}\\&{{P}_{20}}=\frac{100\left(2.310204476\right)}{\left(0.005\right)}\approx\$46204.09 \\\end{align}[/latex]

The account will grow to $46,204.09 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 $46,204.09 – $24,000 = $22,204.09.

Also note that we did round a little early. While [latex] \frac{0.06}{12}[/latex] is exactly equal to 0.005, [latex] \left(1.005\right)^{240}\approx 3.310204476[/latex]. Since we kept every decimal that we saw in our calculator, we did manage to still come up with the correct answer. However, if we had shortened our approximation by even three digits, our final result would have been off. In this example, rounding was used only to highlight the steps necessary if we were unable to enter everything into our calculator at one time.