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!
On
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
$$FV(1)=5000$$ $$FV(2) = 5000(1+e^{0.02})$$ $$...$$
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 $e^{.02t}$.