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kcruz99

  • one year ago

A geometric sequence is defined by the explicit formula a_n=5(-4)^n-1. what is the recursive formula for the nth term of this sequence? A- an=-4a_n-1 B-a_n+1=4a_n C- a_n+1=5a_n D- a_n=-5a_n-1 whenever there is an underscore(_) it means that the letter should be down, if that makes sense lol. Will medal and fan!

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  1. kcruz99
    • one year ago
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    please help!!!!

  2. freckles
    • one year ago
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    \[a_n=5(4)^{n-1} \\ a_{n+1}=5(4)^{n} \\ \text{ also notice } a_n=5(4)^n(4)^{-1} \\ \text{ do you notice what you can replace } 5(4)^n \text{ with ?}\]

  3. freckles
    • one year ago
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    and I notice that one number is -4 so replace my 4's with -4's

  4. anonymous
    • one year ago
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    \[a _{n}=5\left( -4 \right)^{n-1}\] \[a _{n-1}=5\left( -4 \right)^{n-1-1}=5\left( -4 \right)^{n-1}\left( -4 \right)^{-1}=\frac{ 1 }{ -4 } a {n}\] \[a _{n}=-4a _{n-1}\]

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