questionSuppose $24,000 is deposited into an account paying 7.25% interest, which is compoundedcontinuouslyHow much money will be in the account after ten years if no withdrawals or additional depositsare made?

Respuesta :

This is a compound interest question and we have been given:

Principal (P) = $24000

Rate (r) = 7.25%

Years (t) = 10

However, we are told this value is compounded continuously. This means that for every infinitesimal time period, the value keeps being compounded.

The formula for finding the compound interest is:

[tex]\text{Amount}=P(1+\frac{r}{n})^{nt}[/tex]

But because the compounding period is continuous and therefore, infinitesimal,

[tex]\begin{gathered} Amount=P(1+\frac{r}{n})^{nt} \\ But, \\ n\to\infty \\ \\ \therefore Amount=\lim _{n\to\infty}P(1+\frac{r}{n})^{nt} \end{gathered}[/tex]

This is similar to the general formula for Euler's number (e) which is:

[tex]e=\lim _{n\to\infty}(1+\frac{1}{n})^n[/tex]

Thus, we can re-write the Amount formula in terms of e:

[tex]\begin{gathered} \text{Amount}=\lim _{n\to\infty}P(1+\frac{r}{n})^{nt} \\ \text{This can be re-written as:} \\ \\ Amount=\lim _{n\to\infty}P(1+\frac{r}{n})^{\frac{n}{r}\times r\times t}\text{ (move P out of the limit because it is a constant)} \\ \\ \text{Amount}=P\lim _{n\to\infty}((1+\frac{r}{n})^{\frac{n}{r}})^{r\times t} \\ \\ \text{Amount}=P(\lim _{n\to\infty}(1+\frac{r}{n})^{\frac{n}{5}})^{rt} \\ \\ \text{but,} \\ e=(\lim _{n\to\infty}(1+\frac{r}{n})^{\frac{n}{r}} \\ \\ \therefore\text{Amount}=Pe^{rt} \end{gathered}[/tex]

Therefore, we can find the amount of money in the account after 10 years:

[tex]\begin{gathered} \text{Amount}=Pe^{rt} \\ P=24000 \\ r=7.25\text{ \%=}\frac{7.25}{100}=0.0725 \\ t=10\text{ years} \\ \\ \therefore\text{Amount}=24000\times e^{10\times0.0725} \\ \\ \text{Amount}=24000\times2.06473 \\ \\ \therefore\text{Amount}=49553.546\approx49553.55 \end{gathered}[/tex]

Therefore the amount after compounding continuously for 10 years is:

$49553.55