Peter borrows a loan of $200000 from a bank at an interest rate of 6% per annum, compounded monthly. For each successive month after the day when the loan is
taken, loan interest is calculated and then a monthly instalment of $x is
immediately paid to the bank until the loan is fully repaid (the last instalment may
be less $x), where x<200000.
Prove that if Peter has not yet fully repaid the loan after paying the nth instalment, he still owes the bank ${200000(1.005)^(n) - 200x[(1.005)^(n) -1]}.
但是我計到的是$[200000(1.005)^(n) -nx]。
我想問的是，
我的答案是以不變的還款額($x)來計算，
而正確答案的$x應該是逐漸減少，
為甚麽要用正確答案的計法去計算？
還是這些是經濟科的計算=_=？
請幫忙解答，謝謝！
F.5 Maths Qs About borrow loan
2015-02-17 9:27 am
回答 (2)
2015-02-17 6:26 pm
✔ 最佳答案
6% per annum = 0.5% per monthAfter paying the 1st installment, he still owes 200000(1.005) - x ;After paying the 2nd installment, he still owes
[200000(1.005) - x](1.005) - x
= 200000(1.005)^2 - x(1.005) - xAfter paying the 3rd installment, he still owes[200000(1.005)^2 - x(1.005) - x](1.005) - x= 200000(1.005)^3 - x(1.005)^2 - x(1.005) - x⋯⋯So, after paying the nth installment, he still owes200000(1.005)^n - x(1.005)^(n-1) - x(1.005)^(n-2) - x(1.005)^(n-3) - ... - x= 200000(1.005)^n - [x + x(1.005) + x(1.005)^2 ... + x(1.005)^(n-1)]= 200000(1.005)^n - x[(1.005)^n - 1]/(1.005 - 1)= 200000(1.005)^n - 200x[(1.005)^n - 1]
Therefore, he still owes ${200000(1.005)^n - 200x[(1.005)^n - 1]}
2015-02-17 12:29:37 補充：
假設每個月還$10000，則還第一次後尚欠
$200000(1.005) - 10000 = $191000
之後再計算利息，當然是基於$191000，冇可能再基於$200000，因為佢都已經冇欠咁多錢。
2015-02-17 10:27 am
In technical terms, you missed the "time value of money".
You pay $x for n times, but the total is not worth $nx.
You pay $x for n times, but the total is not worth $nx.
收錄日期: 2021-04-15 18:17:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150217000051KK00004