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Math Help - Summation of a Series

  1. #1
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    Summation of a Series

    I'm currently working on geometric sequences but I don't get something really simple about it.

    My first question is:

    How does (4^n)/[3^(n-1)] become 4[(4/3)^(n-1)]

    and how does

    (7^n)/[4^(2n-1)] become (4)[(7^n)/(4^2n)].

    Thank you!!!
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  2. #2
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    Hello, Shelley!

    How does . \frac{4^n}{3^{n-1}}\:\text{ become }\:4\left(\frac{4}{3}\right)^{n-1}

    We have: . \frac{4^n}{3^{n-1}} \;=\;\frac{4\cdot4^{n-1}}{3^{n-1}} \;=\;4\cdot\frac{4^{n-1}}{3^{n-1}} \;=\;4\left(\frac{4}{3}\right)^{n-1}



    How does . \frac{7^n}{4^{2n-1}}\:\text{ become }\:4\cdot\frac{7^n}{4^{2n}}

    We have: . \frac{7^n}{4^{2n-1}} \;=\;\frac{7^n}{4^{2n}\cdot4^{-1}} \;=\;\frac{7^n}{4^{2n}\cdot\frac{1}{4}} \;=\;4\cdot\frac{7^n}{4^{2n}}


    Ha! . . . We can push it even further . . .

    We have: . 4\cdot\frac{7^n}{(4^2)^n} \;=\;4\cdot\frac{7^n}{16^n} \;=\;7\left(\frac{7}{16}\right)^n<br />

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  3. #3
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    thank you so much!!!! however, i am still confused about one part, how does the 4^n become (4)(4^(n-1))?

    thanks again!!
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  4. #4
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    Hello,

    This is because a^{b+c} = a^b a^c

    Hence as n=(n-1)+1, 4^n = 4^{(n-1)+1} = 4^1 4^{n-1} = 4 \times 4^{n-1}
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