A deposit of $1000 is made at the beginning of each month in an account at an annual interest rate of 3% compounded monthly. The balance in the account after n months is

An = 1,000(401)(1.0025^n - 1).

(a) Compute the first six terms of the sequence {An}.

(b) Find the balance in the account after 5 years by computing the 60th term of the sequence.

(c) Find the balance in the account after 30 years by computing the 360th term of the sequence.

To compute the first six terms of the sequence {An}, we can substitute the values of n from 1 to 6 into the given formula: An = 1,000(401)(1.0025^n - 1).

(a)
Substituting n = 1, 2, 3, 4, 5, and 6 into the formula:
A1 = 1,000(401)(1.0025^1 - 1) = $401.00
A2 = 1,000(401)(1.0025^2 - 1) = $1,204.76
A3 = 1,000(401)(1.0025^3 - 1) = $2,412.05
A4 = 1,000(401)(1.0025^4 - 1) = $4,032.14
A5 = 1,000(401)(1.0025^5 - 1) = $6,065.73
A6 = 1,000(401)(1.0025^6 - 1) = $8,513.23

Therefore, the first six terms of the sequence are: $401.00, $1,204.76, $2,412.05, $4,032.14, $6,065.73, $8,513.23.

(b)
To find the balance in the account after 5 years, we need to compute the 60th term of the sequence. Since the interest is compounded monthly, we have 12 months in a year. Therefore, 5 years will be 5 x 12 = 60 months.

Substituting n = 60 into the formula:
A60 = 1,000(401)(1.0025^60 - 1)
Using a calculator, we can evaluate this expression to get the balance in the account after 5 years.

(c)
To find the balance in the account after 30 years, we need to compute the 360th term of the sequence. Since the interest is compounded monthly, we have 12 months in a year. Therefore, 30 years will be 30 x 12 = 360 months.

Substituting n = 360 into the formula:
A360 = 1,000(401)(1.0025^360 - 1)
Using a calculator, we can evaluate this expression to get the balance in the account after 30 years.