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Math Help - Difficult interest rate problem

  1. #1
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    Difficult interest rate problem

    I am not completely sure this involves calculus but it has been presented in our calculus class and I have tried looking at this from all different ways.

    On June 9, 2007 you receive $700 as a graduation present from your wealthy Aunt Thrifty. She will give you $700 more each year that you leave it invested. You leave it invested until June 9, 2054 at the annual interest rate of 3.25%, compounded yearly. How much money is this?

    I know you can do this year by year, but I need a formula. I can't figure out how to include the additional $700 she receives each year and the interest for it.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by cubs3205 View Post
    I am not completely sure this involves calculus but it has been presented in our calculus class and I have tried looking at this from all different ways.

    On June 9, 2007 you receive $700 as a graduation present from your wealthy Aunt Thrifty. She will give you $700 more each year that you leave it invested. You leave it invested until June 9, 2054 at the annual interest rate of 3.25%, compounded yearly. How much money is this?

    I know you can do this year by year, but I need a formula. I can't figure out how to include the additional $700 she receives each year and the interest for it.
    First year: 700 at 3.25% = 1.0325(700)=$722.75
    Second year: (700 at 3.25% + 700) at 3.25% = (1.0325)^2(700) + 1.0325(700)
    Third year: [(700 at 3.25% + 700) at 3.25% + 700] at 3.25% = (1.0325)^3(700) + (1.0325)^2(700) + 1.0325(700)

    N _{th} year: \sum^n_{k=1} 700(1.0325)^k

    From this we can do the total sum for 47 years as:

    \sum^{47}_{k=1} 700(1.0325)^k=\frac{700(1-1.0325^{47})}{1-1.0325}
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  3. #3
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    In the above response the reasoning is correct.
    However, the last seems to be a bit off. See the below: r=1.0325
    Attached Thumbnails Attached Thumbnails Difficult interest rate problem-diff_intrst.gif  
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  4. #4
    Senior Member ecMathGeek's Avatar
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    I will derive the answer (to see if what I wrote was mistaken).

    Let

    S_{47}=\sum^{47}_{k=1} 700(1.0325)^k

    Then

    S_{47}=700(1.0325) + 700(1.0325)^2 +...+700(1.0325)^{47}

    -1.0325S_{47}=-700(1.0325)^2-700(1.0325)^3-...-700(1.0325)^{47}-700(1.0325)^{48}

    Adding these two equations, we get:

    S_{47}-1.0325S_{47}=700(1.0325)-700(1.0325)^{48}=700\cdot1.0325(1-1.0325^{47})

    And so:

    S_{47}=(700\cdot 1.0325)\frac{1-1.0325^{47}}{1-1.0325}

    Thank you, Plato, for checking my work. I went off of memory (and apparently my memory is faulty). Next time I do a problem like this, I will just derive the formula so as to be less likely to make mistakes.
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  5. #5
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    wow thanks. It works out well then.
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