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Math Help - Urgent! AP/GP

  1. #1
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    Urgent! AP/GP

    Hey I need a solution for part (b) only:

    A bank has an account for investors. Interest is added to the account at the end of each year at a fixed rate of 5% of the amount in the account at the beginning of that year. A man and a woman both invest money.

    (b) The woman decides that, to assist her in her everyday expenses, she will withdraw interest as soon as it has been added. She invests $y at the beginning of each year. Show that, at the end of n years, she will have received a total of $[n(n+1)y]/40 in interest.

    Thanks if you could help, I'm having my math paper in about 8 hours' time!
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  2. #2
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    Hello, Margaritas!

    At a bank interest is added to the account at the end of each year
    at a fixed rate of 5% of the amount in the account at the beginning of that year.

    (b) A woman decides that she will withdraw interest as soon as it has been added.
    She invests $y at the beginning of each year.
    Show that, at the end of n years, she will have received a total of $[n(n+1)y]/40 in interest.

    At the end of year-1, she gets 5% of $y = 0.05y = y/20 dollars in interest.

    She invests $y more; at the end of year-2, she gets: 0.05(2y) = 2y/20 dollars in interest.

    She invests $y more; at the end of year-3, she gets: 0.05(3y) = 3y/20 dollars in interest.

    . . . and so on.


    By the end of the nth year, she gets a total of:
    . . y/20 + 2y/20 + 3y/20 + . . . ny/20 dollars in interest.

    . . So she has: .T .= .y(1 + 2 + 3 + ... + n)/20 dollars.

    The sum of the first n integers is: .1 + 2 + 3 + ... + n .= .n(n+1)/2

    . . Therefore: .T .= .y[n(n+1)/2]/20 .= .n(n+1)y/40 dollars.

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  3. #3
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    Thanks Soroban, I get it now!
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