YTC 發問於 科學及數學數學 · 6 年前

F.5 Maths Qs About borrow loan

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!

更新:

RE 邊位都好 & 知足常樂:

因為在五年後回看,五年前的$10在現在可能已是$20,

所以after paying the nth installment, he still owes

200000(1.005)^n - x(1.005)^(n-1) - x(1.005)^(n-2) - x(1.005)^(n-3) - ... - x

要減去$x在還錢那年的價值。

但是,

After paying the 2nd installment, he still owes

[200000(1.005) - x](1.005) - x

為甚麼要以還了第一次錢所剩餘的錢乘以年利率?

2 個解答

評分
  • 6 年前
    最愛解答

    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。

  • 6 年前

    In technical terms, you missed the "time value of money".

    You pay $x for n times, but the total is not worth $nx.

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