In this example:

• [latex]r = 0.06[/latex] ([latex]6\%[/latex])
• [latex]n = 12[/latex] ([latex]12[/latex] compounds/deposits per year)
• [latex]d = $100[/latex] (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_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 [latex]m[/latex] 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 [latex]m[/latex] months, the first deposit will have earned compound interest for [latex]m-1[/latex] months. The second deposit will have earned interest for [latex]m-2[/latex] months. The last month’s deposit ([latex]L[/latex]) 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 [latex]1.005[/latex]:

[latex]1.005{{P}_{m}}=[/latex]

[latex]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 [latex]P_m[/latex]

[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 [latex]m[/latex] months with [latex]12t[/latex], where [latex]t[/latex] is measured in years, gives

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

Recall [latex]0.005[/latex] was [latex]\frac{r}{n}[/latex] and [latex]100[/latex] was the deposit [latex]d[/latex]. [latex]12[/latex] was [latex]n[/latex], the number of deposit each year.

In this example,

[latex]d = $100[/latex]	the monthly deposit
[latex]r = 0.06[/latex]	[latex]6\%[/latex] annual rate
[latex]n = 12[/latex]	since we’re doing monthly deposits, we’ll compound monthly
[latex]t = 20[/latex]	we want the amount after [latex]20[/latex] years

 

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 [latex]$46,200[/latex] after [latex]20[/latex] years.

Notice that you deposited into the account a total of [latex]$24,000[/latex] ([latex]$100[/latex] a month for [latex]240[/latex] months). The difference between what you end up with and how much you put in is the interest earned. In this case it is

[latex]$46,200 - $24,000 = $22,200[/latex].

This example is explained in detail here.

You can view the transcript for “Saving Annuities” here (opens in new window).

Note: This video uses [latex]k[/latex] for [latex]n[/latex] and [latex]N[/latex] for [latex]t[/latex].

In this example, we’re looking for [latex]d[/latex].

[latex]r = 0.08[/latex]	[latex]8\%[/latex] annual rate
[latex]n = 12[/latex]	since we’re depositing monthly
[latex]t = 30[/latex]	[latex]30[/latex] years
[latex]P_{30} = $200,000[/latex]	The amount we want to have in [latex]30[/latex] years

 

In this case, we’re going to have to set up the equation, and solve for [latex]d[/latex].

[latex]\begin{align}&200,000=\frac{d\left({{\left(1+\frac{0.08}{12}\right)}^{30(12)}}-1\right)}{\left(\frac{0.08}{12}\right)}\\&200,000=\frac{d\left({{\left(1.00667\right)}^{360}}-1\right)}{\left(0.00667\right)}\\&200,000=d(1491.57)\\&d=\frac{200,000}{1491.57}=\$134.09 \\\end{align}[/latex]

So you would need to deposit [latex]$134.09[/latex] each month to have [latex]$200,000[/latex] in [latex]30[/latex] years if your account earns [latex]8\%[/latex] interest.

View the solving of this problem in the following video.

You can view the transcript for “Savings annuities – solving for the deposit” here (opens in new window).

Note: This video uses [latex]k[/latex] for [latex]n[/latex] and [latex]N[/latex] for [latex]t[/latex].

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

[latex]d = $100[/latex]	the monthly deposit
[latex]r = 0.03[/latex]	[latex]3\%[/latex] annual rate
[latex]n = 12[/latex]	since we’re doing monthly deposits, we’ll compound monthly

 

We don’t know [latex]t[/latex], but we want[latex]P_t[/latex] to be [latex]$10,000[/latex].

Putting this into the equation:

[latex]10,000=\frac{100\left({{\left(1+\frac{0.03}{12}\right)}^{t(12)}}-1\right)}{\left(\frac{0.03}{12}\right)}[/latex]

Simplifying the fractions a bit

[latex]10,000=\frac{100\left({{\left(1.0025\right)}^{12t}}-1\right)}{0.0025}[/latex]

We want to isolate the exponential term, [latex]1.0025^{12t}[/latex], so multiply both sides by [latex]0.0025[/latex]

[latex]\begin{array}{l} 25 = 100\left({\left(1.0025\right)}^{12t} - 1\right) & \text{Divide both sides by } 100 \\ 0.25 = {\left(1.0025\right)}^{12t} - 1 & \text{Add } 1 \text{ to both sides} \\ 1.25 = {\left(1.0025\right)}^{12t} & \text{Now take the log of both sides} \\ \log(1.25) = \log\left({\left(1.0025\right)}^{12t}\right) & \text{Use the exponent property of logs} \\ \log(1.25) = 12t\log(1.0025) & \text{Divide by } 12\log(1.0025) \\ \frac{\log(1.25)}{12\log(1.0025)} = t & \text{Approximating to a decimal} \\ t = 7.447 \\ \end{array}[/latex]

It will take about [latex]7.447[/latex] years to grow the account to [latex]$10,000[/latex].

This example is demonstrated here:

You can view the transcript for “Use logs to find the time it takes an annuity to grow” here (opens in new window).

Note: This video uses [latex]k[/latex] for [latex]n[/latex] and [latex]N[/latex] for [latex]t[/latex].