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Question 17

If [tex]$6,600 is invested at an interest rate of 6 percent per year, find the value of the investment at the end of 5 years for the following compounding frequencies:

(a) Annually: $[/tex]_____
(b) Semiannually: [tex]$_____
(c) Monthly: $[/tex]_____
(d) Daily: [tex]$_____
(e) Continuously: $[/tex]_____


Sagot :

To find the value of an investment after 5 years for various compounding frequencies, given a principal amount of [tex]$6600 and an annual interest rate of 6%, we can use different formulas for each compounding frequency. ### (a) Annually: For annual compounding, the formula we use is: \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where: - \( P \) is the principal amount ($[/tex]6600).
- [tex]\( r \)[/tex] is the annual interest rate (0.06).
- [tex]\( n \)[/tex] is the number of times interest is compounded per year (1 for annually).
- [tex]\( t \)[/tex] is the time in years (5).

Plugging in the values:

[tex]\[ A = 6600 \left(1 + \frac{0.06}{1}\right)^{1 \times 5} \][/tex]
[tex]\[ A = 6600 (1 + 0.06)^5 \][/tex]
[tex]\[ A = 6600 (1.06)^5 \][/tex]
[tex]\[ A = 8832.288812 \][/tex]

So, the value of the investment with annual compounding after 5 years is approximately [tex]$8832.29. ### (b) Semiannually: For semiannual compounding, we use the same formula but with \( n = 2 \): \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where \( n \) is now 2. \[ A = 6600 \left(1 + \frac{0.06}{2}\right)^{2 \times 5} \] \[ A = 6600 (1 + 0.03)^{10} \] \[ A = 6600 (1.03)^{10} \] \[ A = 8869.848104 \] So, the value of the investment with semiannual compounding after 5 years is approximately $[/tex]8869.85.

### (c) Monthly:
For monthly compounding, we set [tex]\( n = 12 \)[/tex]:

[tex]\[ A = P \left(1 + \frac{r}{n}\right)^{nt} \][/tex]

where [tex]\( n \)[/tex] is now 12.

[tex]\[ A = 6600 \left(1 + \frac{0.06}{12}\right)^{12 \times 5} \][/tex]
[tex]\[ A = 6600 (1 + 0.005)^{60} \][/tex]
[tex]\[ A = 6600 (1.005)^{60} \][/tex]
[tex]\[ A = 8902.411007 \][/tex]

So, the value of the investment with monthly compounding after 5 years is approximately [tex]$8902.41. ### (d) Daily: For daily compounding, we set \( n = 365 \): \[ A = P \left(1 + \frac{r}{n}\right)^{nt} \] where \( n \) is now 365. \[ A = 6600 \left(1 + \frac{0.06}{365}\right)^{365 \times 5} \] \[ A = 6600 (1 + 0.00016438356)^{1825} \] \[ A = 6600 (1.00016438356)^{1825} \] \[ A = 8908.848481 \] So, the value of the investment with daily compounding after 5 years is approximately $[/tex]8908.85.

### (e) Continuously:
For continuous compounding, the formula used is:

[tex]\[ A = Pe^{rt} \][/tex]

where:
- [tex]\( e \)[/tex] is the base of the natural logarithm (approximately equal to 2.71828).
- [tex]\( r \)[/tex] and [tex]\( t \)[/tex] are as defined above.

Plugging in the values:

[tex]\[ A = 6600 \cdot e^{0.06 \times 5} \][/tex]
[tex]\[ A = 6600 \cdot e^{0.30} \][/tex]
[tex]\[ A = 6600 \cdot 1.349858807576003 \][/tex]
[tex]\[ A = 8909.06813 \][/tex]

So, the value of the investment with continuous compounding after 5 years is approximately [tex]$8909.07. In summary, the investment's values at the end of 5 years for different compounding frequencies are: - Annually: $[/tex]8832.29
- Semiannually: [tex]$8869.85 - Monthly: $[/tex]8902.41
- Daily: [tex]$8908.85 - Continuously: $[/tex]8909.07