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Math Help - recurrence relation

  1. #1
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    recurrence relation

    So I understand how to solve it the questions but I think my knowledge of exponents is causing me problems:

    Determine if the sequence {a(n)} is a solution of the recurrence relation a(n) = -3*a(n-1) + 4*a(n-2) if a(n) = 2*(-4)^n+3

    I started as below substituting in the formula:

    a(n) = -3( 2(-4)^(n-1)+3) + 4( 2(-4)^(n-2) + 3)
    = ( -3 *2* -4^(n-1) -9) + 2 * -4^(n-1) + 12

    However,
    I am having trouble with the first term. Any help would be appreciated.
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  2. #2
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    Quote Originally Posted by othnin View Post
    So I understand how to solve it the questions but I think my knowledge of exponents is causing me problems:

    Determine if the sequence {a(n)} is a solution of the recurrence relation a(n) = -3*a(n-1) + 4*a(n-2) if a(n) = 2*(-4)^n+3

    I started as below substituting in the formula:

    a(n) = -3( 2(-4)^(n-1)+3) + 4( 2(-4)^(n-2) + 3)
    = ( -3 *2* -4^(n-1) -9) + 2 * -4^(n-1) + 12

    However,
    I am having trouble with the first term. Any help would be appreciated.


    a_{n} =2(-4)^3(-4)^n =-128(-4)^n

    a_{n-1} = 2(-4)^2(-4)^n = 32(-4)^n

    a_{n-2} = 2(-4)(-4)^n = -8(-4)^n

    hence a_{n} =-3a_{n-1} + 4a_{n-2}
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  3. #3
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    Hello, othnin!

    You started off fine . . .


    Determine if a(n) \:=\:2\!\cdot\!(\text{-}4)^{n+3} is a solution of the recurrence relation:

    . . a(n) \:=\:-3\!\cdot\!a(n-1) + 4\!\cdot\!a(n-2)

    Substitute: . a(n) \;=\;-3\bigg[2(\text{-}4)^{n+2}\bigg] + 4\bigg[2(\text{-}4)^{n+1}\bigg]

    . . . . . . . . . . . . = \;-6(\text{-}4)^{n+1} + 8(\text{-}4)^{n+1}

    . . . . . . . . . . . . = \;2\cdot(\text{-}4)^{n+1}\cdot\bigg[-3(\text{-}4) + 4\bigg]

    . . . . . . . . . . . . = \;2\cdot(\text{-}4)^{n+1}\cdot\bigg[12 + 4\bigg]

    . . . . . . . . . . . . = \;2\cdot(\text{-}4)^{n+1}\cdot16

    . . . . . . . . . . . . = \;2\cdot(\text{-}4)^{n+1}\cdot(\text{-}4)^2

    . . . . . . . . . . . . = \;2\cdot(\text{-}4)^{n+3}\qquad\hdots It's true!

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