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}}=\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]

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.