arithmetic &geometric sequence

sky21212111

2009-11-23 5:55 am

Suppose tom and sally deposit $K into the bank at the end of each year for 20
years. We want to find K such that , at an interest rate of 8% per annum,
compounded annually, the series of payments will have a present value of $1
million.

(a) let U(n) denote the present value of the payment at the end of the n-th year,
i.e. a value which , when deposited into the bank now, will accumulate to $K
bythe end of the n-th year.Show that

(b) To repay the debt by the end of the 20th year, we should have

U(1)+U(2)+....+U(20) = 1000000

Using this and the result of (a) , find K correct to the nearest cent.

thx a lot
long step

回答 (1)

用戶9112

2009-11-23 7:41 am

✔ 最佳答案

The question has ambiguity about timing:
Assume now is start of a year.
(a) The present value = U(n)
The accumulated value at the end of the n-th year=U(n) * (1 + 8%)^n = $K
U(n) = $K / 1.08^n
(b) $K/1.08 + $K/1.08^2 + ... + $K/1.08^20 = $1,000,000
$K/1.08(1 + 1/1.08 + 1/1.08^2 + ... + 1.08^19) = $1000000
(K/1.08)(1 - 1/1.08^20)/(1 - 1/1.08) = 1000000
K = 1080000(1 - 1/1.08)/(1 - 1/1.08^20)
K = 101,852.21
Assume now is end of a year.
(a) The present value = U(n)
The accumulated value at the end of the n-th year=U(n) * (1 + 8%)^(n-1) = $K
U(n) = $K / 1.08^(n-1)
(b) $K + $K/1.08 + ... + $K/1.08^19 = $1,000,000
$K(1 + 1/1.08 + 1/1.08^2 + ... + 1.08^19) = $1000000
(K)(1 - 1/1.08^20)/(1 - 1/1.08) = 1000000
K = 1000000(1 - 1/1.08)/(1 - 1/1.08^20)
K = 94,307.60

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