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I'm a bit stuck on this problem, tried to solve it, but I don't know what is wrong with my way of thinking.

A bank offers a deposit with the interest rate of 5% per annum with quarterly interest capitalization. We make a systematic investment for 5 years by paying at the end of each quarter $100. What is the future (in 5 years) value of this investment?

I tried to do this with following:
$x_1 = PV(1 + 0.05/4)^{4*3/12}+100 = PV(1 + 0.05/4)^{1}+100 $
$x_2 = x_1(1 + 0.05/4)^{2}+100 $
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$x_n =x_{n-1}(1 + 0.05/4)^{n}+100 $
But I don't think it's the correct solution. I tried to simulate this and with $x_0 = 1$, after 20 iterations (n = 20 = 4 capitalization for 5 years, right?), $x_20 = 13658,31$ which looks unreal. I wish I know how to do it.

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1 Answers1

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Your terms are not so bad. Lets start with the first payment.

$x_0=100$

This payment is compounded with the (quarterly) interest rate. And additionally $100 are payed in.

$i_4=\frac{0.05}{4}$, $q_4=1+\frac{0.05}{4}$

$x_1=x_0\cdot q_4 +100=100\cdot q_4+100$

Then the next quarter.

$x_2=x_1\cdot q_4 +100=(100\cdot q_4+100)\cdot q_4+100=100q_4^2+100q_4+100$

Thus $x_n=100q_4^n+100q_4^{n-1}+\ldots+100q_4+100=100\cdot(q_4^n+q_4^{n-1}+\ldots+q+1)$

This expression can be written in a closed form by using the formula for the nth sum of the geometric series.

$x_n=100\cdot \frac{q_4^{n+1}-1}{q_4-1}=100\cdot \frac{q_4^{n+1}-1}{i_4}$

REMARK to your post

$x_n =x_{n-1}(1 + 0.05/4)^{n}+100 $ But I don't think it's the correct solution.

It is just $x_n =x_{n-1}(1 + 0.05/4)+100 $-without the exponent n. You get to the next quarter by multiplying the amount of money of the previous month by $q_4$ and adding $100.

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