Question

How much should I have to save each month if I’d wanted to end up with $ 5000 at the end of 5 years, assuming that I earn interest 4% compound monthly ? How much did i put in total ? How much interest did i earn?

Answer 1

Consider how the amount grows. Let the regular amount saved per month be `a` dollars, and let the compounding factor be `I` - in this case `I=1+(4%)/100=1.04`.

After one month, you have `Ia`.
After two months, you have `I(a+Ia)=a(I+I^2)`.
After three months, you have `I(a+I(a+Ia))=a(I+I^2+I^3)`.

After `n` months, you have `a(I+I^2+I^3+...+I^n)=asum_(j=1)^nI^j`.

This is a geometric series, which can be summed with the formula
`a((1-r^n)/(1-r))`, where `r` is the ratio of successive terms (`r=I` here), and `a` is the starting value - note that the formula includes the term with `I^0`, not included here, so we need to subtract off `a`:
`a((1-r^n)/(1-r))-a`
`a((1-r^n)/(1-r)-1)`
`a(((1-r^n)-(1-r))/(1-r))`
`a((r-r^n)/(1-r))`
`ar((1-r^(n-1))/(1-r))`
`aI((1-I^(n-1))/(1-I))`

Wikipedia gives the proof of this formula:
https://en.wikipedia.org/wiki/Geometric_series#Formula

After 5 years, `n=12*5=60`, so for our particular problem, where we set the total value to $5000, we obtain the formula

`a*1.04*((1-1.04^59)/(1-1.04))=5000`, which we can solve for `a` to deduce how much you should save each month.

`1.04^59=10.1150263539...`
`1-1.04^59=-9.1150263539...`
`1-1.04=-0.04`
`a=(5000*(-0.04))/(1.04*(-9.1150263539...))`
`a=21.0978756222...`

So to have $5000 after 5 years, in an account compounding interest monthly at 4%, you would need to save each month $21.10. If you ever find an account that pays this much interest in real life, grab it and never let go!