Question

if 6000 dollars is invested in a bank account at an interest rate of 6 per cent per year,
find the amount in the bank after 6 years if interest is compounded annually:

find the amount in the bank after 6 years if interest is compounded quaterly:

find the amount in the bank after 6 years if interest is compounded monthly:

finally, find the amount in the bank after 6 years if interest is compounded continuously:

Explanation:

Step1: Recall the compound - interest formula

The compound - interest formula is \(A = P(1+\frac{r}{n})^{nt}\), where \(P\) is the principal amount (\(P = 6000\)), \(r\) is the annual interest rate (as a decimal, \(r=0.06\)), \(t\) is the number of years (\(t = 6\)), and \(n\) is the number of times interest is compounded per year. For continuous compounding, the formula is \(A=Pe^{rt}\).

Step2: Calculate when compounded annually (\(n = 1\))

Substitute \(P = 6000\), \(r = 0.06\), \(t = 6\), \(n = 1\) into \(A = P(1+\frac{r}{n})^{nt}\).

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Using a calculator, \((1.06)^{6}\approx1.418519\), so \(A = 6000\times1.418519=8511.11\)

Step3: Calculate when compounded quarterly (\(n = 4\))

Substitute \(P = 6000\), \(r = 0.06\), \(t = 6\), \(n = 4\) into \(A = P(1+\frac{r}{n})^{nt}\).

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Using a calculator, \((1.015)^{24}\approx1.43076878\), so \(A=6000\times1.43076878 = 8584.61\)

Step4: Calculate when compounded monthly (\(n = 12\))

Substitute \(P = 6000\), \(r = 0.06\), \(t = 6\), \(n = 12\) into \(A = P(1+\frac{r}{n})^{nt}\).

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Using a calculator, \((1.005)^{72}\approx1.4320449\), so \(A=6000\times1.4320449=8592.27\)

Step5: Calculate when compounded continuously

Substitute \(P = 6000\), \(r = 0.06\), \(t = 6\) into \(A=Pe^{rt}\).

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Using a calculator, \(e^{0.36}\approx1.4333294\), so \(A = 6000\times1.4333294=8599.98\)

Answer:

When compounded annually: \(8511.11\)
When compounded quarterly: \(8584.61\)
When compounded monthly: \(8592.27\)
When compounded continuously: \(8599.98\)