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Math Help - sequence and series

  1. #1
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    sequence and series

    write the formula for the nth term of the sequence
    -(1/1), (3/2), -(7/6), (15/24), -(31/120)
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  2. #2
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    (-1)^n \frac{2^n-1}{n!}
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  3. #3
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    a_1=-\frac{1}{1},...,a_n=(-1)^n\frac{b_n}{c_n}=(-1)^n\frac{b_{n-1} +2^{n-1} }{nc_{n-1}}
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  4. #4
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    Hello, bekah!

    Julius is correct!


    Write the formula for the n^{th} term of the sequence:
    . . -\frac{1}{1},\;\frac{3}{2},\; -\frac{7}{6},\;\frac{15}{24},\; -\frac{31}{120},\;\hdots

    The signs alternate.
    . . This involves: . {\color{blue}(\text{-}1)^n}



    The numerators are: . 1,\; 3,\; 7,\; 15,\; 31, \hdots

    You might recognize them: one less than a power-of-2.
    . . \begin{array}{ccc}<br />
1 &=& 2^1 - 1 \\ 3 &=& 2^2-1 \\ 7 &=& 2^3 - 1 \\ 15 &=& 2^4-1 \\ 31 &=& 2^5-1 \\ \vdots && \vdots \end{array}

    Hence, the numerator is: . {\color{blue}2^n-1}



    The denominators are: . 1,\;2,\;6,\;24,\;120,\;\hdots

    You might recognize these as factorials:
    . . \begin{array}{ccc}1 &=&1! \\ 2 &=& 2! \\ 6 &=& 3! \\ 24 &=& 4! \\ 120 &=& ! \\ \vdots & \vdots \end{array}

    Hence, the denominator is: . {\color{blue}n!}



    Therefore, the n^{th} term is: . a_n \;=\;(\text{-}1)^n\frac{2^n-1}{n!}

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