$8877.62
This is an example of an investment with compound interest and regular contributions each period. So let's examine it closely.
To make things easier, I'll work with individual months instead of entire years. So we'll have 60 periods (12 * 5) and the interest rate for each period will be 5/12 = 0.41666%. For the interest, we'll add 1 and call the value z, so that our math doesn't have to keep adding. The contribution each month will be called C which will be made at the beginning of the month. So let's look at the end of the first few months.
1. Cz
2. Cz^2 + Cz = C(z+ z^2)
3. Cz^3 + Cz^2 + Cz = C(z + z^2 + z^3)
...
n. Cz^n + Cz^(n-1) + ... + Cz = C(z + z^2 + ... + z^n)
The (z + z^2 + ... + z^n) part is the sum of a geometric series. If you look it up, you'll see that it reduces to this
(z^(n+1) - z)/(z-1)
Now we can calculate how much Michelle will have after 60 months.
T = 130(z^61 - z)/(z-1)
z = 1 + 0.05/12 = 1.004166667
so
T = 130(1.004166667^61 - 1.004166667)/(1.004166667-1)
T = 130(1.288706006 - 1.004166667)/.004166667
T = 130(0.28453934)/.004166667
T = 130(68.28944152)
T = 8877.627398
So Michelle will have after 5 years, $8877.62 saved.
