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I have to find a formula for a problem that involves yearly continuous compounding at 2% with annual contributions of 5000$. We know y(0)=0, so there is initially no money in this account.

Let FV= Future value, n= number of years that have passed,P= initial value of account, i= interest rate in decimals.

Thank you!

Alex.G
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  • What are your thoughts? – IMOPUTFIE Sep 23 '19 at 16:03
  • I know that continuous compounding is calculated with FV=P*e^(ni) , I just don't understand how to take into account the annual contributions – Alex.G Sep 23 '19 at 16:07
  • You just have a geometric progression. $e^{n\iota}=\left(e^\iota\right)^n$ – saulspatz Sep 23 '19 at 16:18
  • How do I turn that into a complete formula that gives me the amount in the account for every year? @saulspatz – Alex.G Sep 23 '19 at 16:23
  • What exactly is the question? – saulspatz Sep 23 '19 at 16:24
  • Suppose you want to start saving for retirement. You decide to continuously invest $5000 of your income each year in a risk-free investment with a 2% yearly interest rate, compounded continuously. Find the equation for y(t) (the amount of money in the account at any given year). @saulspatz – Alex.G Sep 23 '19 at 16:26

2 Answers2

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The expression for the future value of the account in $n$-year is

$$\text{FV}(n)=5000 [e^{0.02(n-1)}+e^{0.02(n-2)}+e^{0.02(n-3)}+ \>...\> e^{0.02}+1]$$

which is a geometric series and can be summed as,

$$\text{FV}(n)=5000 \frac{ e^{0.02n}-1}{e^{0.02}-1}$$

So, you have

$$FV(1)=5000$$ $$FV(2) = 5000(1+e^{0.02})$$ $$...$$

Quanto
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  • It says the answer is wrong. If it helps, we know that after 35 years, there is 253439$ in the account @Quanto – Alex.G Sep 23 '19 at 16:54
  • @Alex.G - you'd get 253439 if you use a similar version $\text{FV}(35)=5000 \frac{ e^{0.02\times 35}-1}{0.02}$. But, then, it is no longer continuous compounding. – Quanto Sep 23 '19 at 17:25
  • That was the correct equation thank you so much! My professor probably didn't write the correct instruction. @Quanto – Alex.G Sep 23 '19 at 17:35
  • @Alex.G - no problem – Quanto Sep 23 '19 at 17:36
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Here's a formula for the amount in the account at the end of $n$ years. There have been $n$ contributions of $5000$ each, which have been in the account earning interest, for $n,n-1,n-2,\dots,1$ years. Their accumulated value is $$5000\sum_{k=1}^ne^{.02k}=5000\sum_{k=1}^n\left(e^{.02}\right)^k=5000{e^{.02(n+1)}-e^{.02}\over e^{.02}-1}$$

To find the value at time $n+t$, where $n$ is a nonnegative integer and $0<t<1$, we have only to multiply by $e^{.02t}$.

saulspatz
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