Question

You invest $500 in an account with an annual interest rate of 1.1%, compounded continuously. How much money is in the account after 10 years? Round your answer to the nearest whole number.

Answer 1

There is a formula to calculate continuous compound, which is A = Qe\[A=Q*e ^{rt}\], here A is the balance you wanna calculate, Q is the principle money, e is the natural exponent, r is the annual interest rate, t is the number of years passed. For your question, you can insert in all the numbers to calculate the result.

To prove the formula, suppose a principle Q is invested in the bank at an annual interest r, compounded once a year, so the balance after the first year is \[A=Q*(1+r)\]. The pattern of multiplying the previous principle by (1+r) is repeated each successive year. Suppose t years has passed, then \[A=Q*(1+r)^{t}\].

To accommodate more frequent (quarterly, monthly, or daily) compounding of
interest, let n be the number of compoundings per year and let t be the number of years.Then the rate per compounding is r/n, and the account balance after years is \[A=Q*(1+r/n)^{nt}\].

For continuous compounding, let m=n/r, then \[A=Q*(1+r/n)^{nt}\] is transferred into \[A=Q *(1+r/mr)^{mrt}\], which is \[A = Q *( (1+1/m)^{m})^{rt}\]. As m increases without bound, \[(1+1/m)^{m}\] approaches e, so we get the formula above.

For the easy way, you can just remember it. Hope this answer your question.