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Math Help - hello new to consumer math need help

  1. #1
    tdmj90
    Guest

    hello new to consumer math need help

    How much can you accumulate in a savings plan if your annual year-end deposit is $400, and you earn 4% annual interest for 12 years? S =[(P(1+r)n 1]/r

    can someone help simplify this for me? thank you
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  2. #2
    Member
    Joined
    Mar 2008
    Posts
    148
    Once way you can do this is to set a table:

    [HTML]<table border = ".1" cellpadding = "4">
    <tr>
    <td> </td><td> Deposited</td><td>Interest Earned</td><td>Total Balance</td>
    </tr>
    <tr>
    <td>End of year 1</td><td>400</td><td>0</td><td>400</td>
    </tr>
    <tr>
    <td>End of year 2</td><td>400</td><td>.04*400 = 16</td><td>400+400+16 = 816</td>
    </tr>
    <tr>
    <td>End of year 3</td><td>400</td><td>.04*816 = 32.64</td><td>816+400+32.64 = 1248.64</td>
    </tr>

    </table>[/HTML]

    and continue in this fashion for 12 years. Or observe that:
    after 1 year you have a balance of 400,

    after 2 years you have a balance of 400+400(1.04) = 400((1.04)^0+(1.04)^1)

    after 3 years your balance is 400+400(1.04)+400(1.04)^2 = 400((1.04)^0 + (1.04)^1+(1.04)^2)

    and after n years you have a balance of 400((1.04)^0 + (1.04)^1 + ... + (1.04)^{n-1})

    The sum of powers of 1.04 is a geometric series which simplifies to the formula: \frac{1-(1.04)^n}{1-1.04}

    So after n years you should have saved: 400\left(\frac{1-(1.04)^n}{1-1.04}\right).
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