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anonymous

  • one year ago

The following function defines a recursive sequence: f(0) = -3 f(1) = 6 f(n) = -2•f(n -1) - f(n - 2); for n > 1 Which of the following sequences is defined by this recursive function?

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  1. anonymous
    • one year ago
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    -3, 6, -9, -12, … -3, 20, -95, 480, … -3, 6, -9, 12, … -3, -20, -95, -480, …

  2. SolomonZelman
    • one year ago
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    I will change my notations to \(a_{n}\) if you don't mind, ok?

  3. anonymous
    • one year ago
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    um okay i guess

  4. SolomonZelman
    • one year ago
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    But you are given the first two terms \(a_0\) and \(a_1\) so just based on that you can exclude the rest of the options

  5. SolomonZelman
    • one year ago
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    u have only 2 possible options after doing elimination, right? and they are A and C

  6. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle a_n=-2\cdot \left(a_{n-1}\right) -\left(a_{n-2}\right) }\) \(\large\color{black}{ \displaystyle a_2=-2\cdot \left(a_{2-1}\right) -\left(a_{2-2}\right) }\) \(\large\color{black}{ \displaystyle a_2=-2\cdot \left(a_{1}\right) -\left(a_{0}\right) }\) \(\large\color{black}{ \displaystyle a_2=-2\cdot \left(6\right) -\left(-3\right) }\) \(\large\color{black}{ \displaystyle a_2=-12 +3 }\) \(\large\color{black}{ \displaystyle a_2=-9}\)

  7. SolomonZelman
    • one year ago
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    then, you can find \(a_3\) using the same formula

  8. anonymous
    • one year ago
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    so its c?

  9. SolomonZelman
    • one year ago
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    let me see.... \(\large\color{black}{ \displaystyle a_3=-2\cdot \left(a_{3-1}\right) -\left(a_{3-2}\right) }\) \(\large\color{black}{ \displaystyle a_2=-2\cdot \left(a_{2}\right) -\left(a_{1}\right)=-2\cdot(-9)-(6)=18-6=12 }\)

  10. SolomonZelman
    • one year ago
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    yes C is right

  11. SolomonZelman
    • one year ago
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    you can deduce that by logic, that you have a negative term \(a_{n-1}\) (in this case a negative term \(a_{2}\) (which is =-9) |--> so when multiplied times -2 it becomes twice as much and positive. This negative term \(a_2\) has a greater absolute value (or greater magnitude) than \(a_{n-2}\) (and \(a_{n-2}\) in this case is \(a_1\) which is 6) So you are having a case \(a_3=-2\times ({\rm -greater})~-~({\rm smaller})~~~~~~~\Rightarrow~\rm positive\)

  12. anonymous
    • one year ago
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    okay great thank you so much for your help! : )

  13. SolomonZelman
    • one year ago
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    and you know that it is 12 and -12, having eliminated every option besides A and C. And since the result (for \(a_2\)) must be positive, therefore it is 12 (not -12), and thus the answer is C.

  14. SolomonZelman
    • one year ago
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    this is kidnd of an implicit.... yw

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