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Math Help - a=3n^(n-1)+2(-1)^n for neN

  1. #1
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    a=3n^(n-1)+2(-1)^n for neN

    I was wondering if anyone could help me with the following question:

    a=3n^(n-1)+2(-1)^n for neN-3.55.png


    Thanks,
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  2. #2
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    Try induction, it's real easy in this case.
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  3. #3
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    Quote Originally Posted by Jose27 View Post
    Try induction, it's real easy in this case.
    ohh ok
    So would i use n=3 as my basecase and the sub k+1 for all n ?
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  4. #4
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    Quote Originally Posted by jess0517 View Post
    ohh ok
    So would i use n=3 as my basecase and the sub k+1 for all n ?
    Basically, just prove the case n=3, then assume it's valid for n=k and, using this, prove that it is valid for k+1.
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  5. #5
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    Hello, jess0517!

    We can prove it by deriving that function ourselves . . .



    \text{Let }\{a\} \text{ be a sequence satisfying:}

    . . a_1 = 1,\;\;a_2 = 8,\;\;a_n \:=\:a_{n-1} + 2a_{n-2}\:\text{ for }n \ge 3.


    \text{Prove that: }\;a_n \;=\;3\!\cdot\!2^{n-1} + 2(\text{-}1)^n\:\text{ for }n \in N.

    We have: . a_n - a_{n-1} - 2a_{n-2} \;=\;0


    Let X^n = a_n

    . . and we have: . X^n - X^{n-1} - 2X^{n-2} \:=\:0


    Divide by X^{n-2}\!:\;\;X^2 - X - 2 \:=\:0

    . . Then: . (X - 2)(X+1) \:=\:0 \quad\Rightarrow\quad X \:=\:2,\:-1


    The generating function is of the form: . f(n) \;=\;A\!\cdot\!2^n + B\!\cdot\!(\text{-}1)^n


    We know the first two values of the sequence:

    . . \begin{array}{cccccccccc}<br />
f(1) = 1\!: && 2A - B &=& 1 & [1] \\<br />
f(2) = 8\!: && 4A + B &=& 8 & [2] \end{array}

    Add [1] and [2]: . 6A \:=\:9 \quad\Rightarrow\quad A \:=\:\frac{3}{2}

    Substitute into [1]: . 2(\frac{3}{2}) - B \:=\:1 \quad\Rightarrow\quad B \:=\:2

    Hence: . f(n) \;=\;\frac{3}{2}\!\cdot\!2^n + 2(\text{-}1)^n


    Therefore: . a_n \;=\;3\!\cdot\!2^{n-1} + 2(\text{-}1)^n

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  6. #6
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    Thread closed due to this member deleting questions after getting help.
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