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Math Help - sum of the first n

  1. #1
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    sum of the first n

    If the sum of the first n terms in the sequence  a_1, a_2, a_3, ... is equal to 1/n, find the product of its first 2007 terms.
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  2. #2
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    Hello, perash!

    I must be missing some simple. .I'm getting no pattern . . .


    If the sum of the first n terms in the sequence: a_1,\:a_2,\:a_3,\:\cdots is equal to \frac{1}{n},
    find the product of its first 2007 terms.

    We are told that: . \begin{array}{ccc}a_1 & = & 1 \\  a_1+a_2 & = & \quad\frac{1}{2} \\ a_1+a_2+a_3 & = & \frac{1}{3} \\ a_1+a_2+a_3+a_4 &=& \quad\frac{1}{4} \\ \vdots & & \vdots\end{array}

    This gives us: . \begin{array}{ccccc}<br />
a_1 &=& 1 & = & \frac{1}{1!} \\  a_2 &=& \quad\text{-}\frac{1}{2} &=& \quad\text{-}\frac{1}{2!} \\ a_3 &=&\frac{5}{6} &=& \frac{5}{3!} \\ a_4 &=& \quad\text{-}\frac{7}{12} &=&\quad\text{-}\frac{14}{4!} \\ a_5 &=& \frac{47}{60} &=&\frac{94}{5!} \\ <br />
a_6 &=& \quad\text{-}\frac{37}{60} &=& \quad\text{-}\frac{444}{6!} \\<br />
\vdots & & & & \vdots<br />
\end{array}


    The denominators are: . 1!,\:2!,\:3!,\,\cdots\:2007!

    The numerators are: . 1,\:\text{-}1,\:5,\:\text{-}14,\:94,\:\text{-}444,\:\cdots
    . . and I see no pattern there . . .


    The desired product is: . P_{2007} \;=\;\frac{(1)(\text{-}1)(5)(\text{-}14)(94)(\text{-}444)\:\cdots} {(1!)(2!)(3!)(4!) \cdots(2007!)}

    . . Good luck!

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  3. #3
    Super Member PaulRS's Avatar
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    \sum_{i=1}^n{a_i}=\frac{1}{n}

    But \sum_{i=1}^{n-1}{a_i}=\frac{1}{n-1}

    Then: a_n=\frac{1}{n}-\frac{1}{n-1}=-\frac{1}{n\cdot{(n-1)}} for n\geq{2}

    So: P_n=\prod_{i=1}^{n}{a_i}=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i\cdot{(i-1)}}}

    P_n=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i\cdot{(i-1)}}}=(-1)^{n-1}\prod_{i=2}^{n}{\frac{1}{i}}\prod_{i=1}^{n-1}{\frac{1}{i}}

    Therefore: P_n=(-1)^{n-1}\cdot{\frac{1}{n!\cdot{(n-1)!}}} when n\geq{2}
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