Respuesta :
Answer:
[tex]\begin{array}{ccl}Year&&End \ of \ year \ balance\\1&&\$4,483.18\\2&& \$3,088.68\\3&&\$1,596.57\\4&&0\end{array}[/tex]
Step-by-step explanation:
The initial amount invested, P = $1,600
Danny's age at which the amount, A, is to be used for his college, t = 19 years
The number of equal annual payments, m, to be withdrawn from the amount = 4
The compound interest on the account, r = 7% = 0.07
Let, A, represent the amount at the year the annual withdrawals starts to be made, we have;
[tex]A = P \times \left(1+\dfrac{r}{n} \right)^{n \times t}[/tex]
n = The number of times the interest is applied annually = 1
Therefore;
[tex]A = 1,600 \times \left(1+\dfrac{0.07}{1} \right)^{1 \times 19} \approx 5,786.444[/tex]
The amount, m, is withdrawn at start of Danny's first year in college to give the amount in the account = A - m
The amount in the account at the end of the first year with compound interest, r = (A - m)×(1 + r)¹ = (A - m)×(1 + r)
At the stat of the second year, the second withdrawal is made to give the starting amount = (A - m)×(1 + r) - m
The amount in the account at the end of the second year = ((A - m)×(1 + r) - m)×(1 + r)
At the start of the third year, the amount in the account =  ((A - m)×(1 + r) - m)×(1 + r) - m
At the end of the third year, we have the amount in the account  = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r)
At the start of the forth year, the last yearly installment is withdrawn from the account and we have 0 balance in the account.
Therefore, on the fourth year, we have the amount in the account = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) - m = 0
(A - m)×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0
A×(1 + r)³ - m×(1 + r)³ - m×(1 + r)² - m × (1 + r) - m = 0
A×(1 + r)³ = m×(1 + r)³ + m×(1 + r)² + m×(1 + r) + m = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)
A×(1 + r)³ = m × ((1 + r)³ + (1 + r)² + (1 + r) + 1)
m = A×(1 + r)³/((1 + r)³ + (1 + r)² + (1 + r) + 1)
∴ m = 5,786.444×(1 + 0.07)³/((1 + 0.07)³ + (1 + 0.07)² + (1 + 0.07) + 1) ≈ 1,596.56
The amount withdrawn annually, m ≈ $1,596.56
The amount in the account at the end of the each year is given as follows;
First year = (A - m)×(1 + r) = (5,786.444 - 1,596.56)×(1 + 0.07) ≈ 4,483.17588
Second year = ((A - m)×(1 + r) - m)×(1 + r) = ((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) = 3,088.68
Third year = (((A - m)×(1 + r) - m)×(1 + r) - m) × (1 + r) = (((5,786.444 - 1,596.56)×(1 + 0.07) - 1,596.56)×(1 + 0.07) - 1,596.56) × (1 + 0.07) = 1,596.57
At the end of the first year, we have $4,483.17588
At the end of the second year, we have $3,088.68
At the end of the third year, we have $1,596.57
At the end of the fourth year, we have 0
