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Math Help - How do I get from here to here? - basic algebra

  1. #1
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    How do I get from here to here? - basic algebra

    Hi!

    I have no idea how to do this because the book I'm reading doesn't show me how:

    A x sum of 1/(1+r)t where t goes on for however long, apparently equals A/r x [1 - 1/(1+r)t ]

    how do you get there?

    I'd be so thankful for someone to show me )
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  2. #2
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    Re: How do I get from here to here? - basic algebra

    Hello, mathbeginner97!

    There is a typo . . .


    Show that: .

    A\sum^n_{t=1} \frac{1}{(1+x)^t} \;=\;\frac{A}{x}\left[1 - \frac{1}{(1+x)^n}\right]

    We have: . S \;=\; \frac{A}{1+x} + \frac{A}{(1+x)^2} + \frac{A}{(1+x)^3} + \cdots + \frac{1}{(1+x)^n}

    This is geometric series with first term a \,=\,\frac{A}{1+x},
    . . common ratio r \,=\,\frac{1}{1+x} and n terms.

    The sum is: . S \;=\;a\,\frac{1-r^n}{1-r}


    S \;=\;\frac{A}{1+x}\left[\frac{1-(\frac{1}{1+x})^n}{1-(\frac{1}{1+x})}\right] \;=\;\frac{A}{1+x}\left[\frac{1 - \frac{1}{(1+x)^n}}{\frac{1+x-1}{1+x}}\right] \;=\;\frac{A}{1+x}\left[\frac{1-\frac{1}{(1+x)^n}}{\frac{x}{1+x}}\right]

    S \;=\;A\left[\frac{1-\frac{1}{1+x)^n}}{x}\right] \;=\;\frac{A}{x}\left[1 - \frac{1}{(1+x)^n}\right]
    Thanks from mathbeginner97
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  3. #3
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    Re: How do I get from here to here? - basic algebra

    Thank you very much!

    I appreciate your help a lot!
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