Math Problem Statement

Suppose $2,000.00 dollars is invested in an account for 13 years. Find the value of the investment at the end of 13 years if the interest is: (a) 5.2% compounded annually: (b) 5.2% compounded semiannually: (c) 5.2% compounded quarterly: (d) 5.2% compounded monthly: (e) 5.2% compounded daily (ignore leap years): and answers

Solution

To calculate the future value of an investment with compound interest, we use the compound interest formula:

A=P(1+rn)ntA = P \left(1 + \frac{r}{n}\right)^{nt}

Where:

  • AA is the future value of the investment
  • PP is the principal amount (initial investment)
  • rr is the annual interest rate (in decimal form)
  • nn is the number of times the interest is compounded per year
  • tt is the time the money is invested for, in years

The principal amount (PP) is $2,000.00, the interest rate (rr) is 5.2% or 0.052 in decimal, and the time (tt) is 13 years.

(a) Compounded annually:

For annual compounding, n=1n = 1.

A=2000(1+0.0521)1×13A = 2000 \left(1 + \frac{0.052}{1}\right)^{1 \times 13}

(b) Compounded semiannually:

For semiannual compounding, n=2n = 2.

A=2000(1+0.0522)2×13A = 2000 \left(1 + \frac{0.052}{2}\right)^{2 \times 13}

(c) Compounded quarterly:

For quarterly compounding, n=4n = 4.

A=2000(1+0.0524)4×13A = 2000 \left(1 + \frac{0.052}{4}\right)^{4 \times 13}

(d) Compounded monthly:

For monthly compounding, n=12n = 12.

A=2000(1+0.05212)12×13A = 2000 \left(1 + \frac{0.052}{12}\right)^{12 \times 13}

(e) Compounded daily (assuming 365 days in a year):

For daily compounding, n=365n = 365.

A=2000(1+0.052365)365×13A = 2000 \left(1 + \frac{0.052}{365}\right)^{365 \times 13}

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Math Problem Analysis

Mathematical Concepts

Compound Interest
Exponential Growth
Financial Mathematics

Formulas

A = P(1 + r/n)^(nt)

Theorems

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Suitable Grade Level

Grades 9-12