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Suppose I have money ($x$) in a bank account bank with compound interest of $5\%$ annually paid monthly. Bank gives me $20,000$ usd a month and the money $x$ finishes in $20$ years ($240$ months). How to calculate how much was $x$?

For example if $x=1,000,000$, after first month the money will reduce to $x'=1,000,000+1,000,000\times 0.05\times \frac{1}{12}-20,000=984,167$.

I know how to formulate it but it's very long messy equation which I have no idea how to solve it.

Thank you

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    $\begingroup$ In other words, an annuity $\endgroup$ Aug 27 '15 at 9:16
  • $\begingroup$ @HenningMakholm - Yes, x will be the money if one to receive it once upfront instead of 240-month 20k-a-month. $\endgroup$
    – L.G.
    Aug 27 '15 at 9:21
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By the way, $20$ years is $240$ months.

I will continue with your formulation. Let $r=1+\frac{0.05}{12}$.

After $1$ month, the bank has $$xr-20000$$

After $2$ month, the bank has $$(xr-20000)r-20000=xr^2-20000r-20000$$

After $3$ months, the bank has $$xr^3-20000r^2-20000r-20000$$

Following the pattern, after $240$ months, the bank has $$xr^{240}-20000(1+r+r^2+\cdots r^{239})\\ =xr^{240}-20000(\frac{r^{240}-1}{r-1})$$

Can you continue from here?

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  • $\begingroup$ Yes, I can! It will be $x=3030506$. Thank you very much. :) $\endgroup$
    – L.G.
    Aug 27 '15 at 9:27
  • $\begingroup$ @LuckyGuy: Right! Will change. $\endgroup$
    – KittyL
    Aug 27 '15 at 9:31
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In practice I always like to balance the equation at a certain time (usually the beginning or the end) when I did a problem about interests while I am still in high school. Let us consider the values at the end of $20$ years. How much does the $k$-th payment worth at the end of $20$ years? Imagine you put it back into deposit right after you get the payment, it should yield $$20000 \times (1+\frac{0.05}{12})^{480-k}$$ and of course it would worth the same or otherwise you can make money by arbitrage. Therefore, this profolio worths totally $$\sum_{k=1}^{480} 20000 \times (1+\frac{0.05}{12})^{480-k} = 20000 \times \frac{(1+\frac{0.05}{12})^{480} - 1}{(1+\frac{0.05}{12}) - 1}$$ at the end of $20$ years. On the other hand, if it worths $V$ now, it would worth $$V \times (1+\frac{0.05}{12})^{480}$$ at the end of $20$ years. Therefore we have the equation $$V \times (1+\frac{0.05}{12})^{480} = 20000 \times \frac{(1+\frac{0.05}{12})^{480} - 1}{(1+\frac{0.05}{12}) - 1}$$

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