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Math Help - rewrite as geometric series

  1. #1
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    rewrite as geometric series

    Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
    Thank you.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by cowboys111 View Post
    Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
    Thank you.
    \sum_{n=0}^{\infty}\frac{4^{n+2}}{7^{n-1}}=\sum_{n=0}^{\infty}\frac{16}{\frac{1}{7}}\bigg  (\frac{4}{7}\bigg)^{n}=112\sum_{n=0}^{\infty}\bigg  (\frac{4}{7}\bigg)^{n}=112\frac{1}{1-\frac{4}{7}}=261.33
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  3. #3
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    Quote Originally Posted by cowboys111 View Post
    Hi, i am trying to rewrite this problem as a geometric series in order to solve it but im having some problems. 4^(k+2)/7^(k-1). from there i get 112(4^k/7^k) which seems fine to me with a=112 and r=(4/7), but my book has a=64 and r=(4/7). Can someone please teel me if i did something wrong or what is happening?
    Thank you.
    I think you're right on your geometric series as 112(4^k/7^k) but if a = 64 (the first value of your geometric series), then your lower limit of summation must be 1. Putting in 1 for k yields a = 64 for your first value.

    Not sure but I think that's what it is.
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