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Math Help - Recursive formula to a power series

  1. #1
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    Recursive formula to a power series

    sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
    sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

    What is the recursive formula for this that connects the nth term with the (n+1)th term?

    Thanks in advance
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Sam1111 View Post
    sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
    sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

    What is the recursive formula for this that connects the nth term with the (n+1)th term?

    Thanks in advance
    \sum_{k=0}^{\infty} \frac{(-1)^{k+1}x^k}{(k+2)!3^{2k}}

    so:

    a_k=\frac{(-1)^{k+1}x^k}{(k+2)!3^{3k}}

    so:

    a_{k+1}=\frac{(-1)^{k+1}(-1)x^kx}{(k+2)!(k+3)3^{3k}3^2}

    CB
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  3. #3
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    Quote Originally Posted by Sam1111 View Post
    sum ((-1)^(k+1)*x^k)/((k+2)!*3^(2*k)) from k =0 to infinity
    sum ((-1)\^{}(k+1)*x\^{}k)/((k+2)!*3\^{}(2*k)) from k =0 to infinity is the series (with my attempt at latex using some converter - comes up with latex error with the math square brackets around it..?)

    What is the recursive formula for this that connects the nth term with the (n+1)th term?

    Thanks in advance

    Let's see: \sum\limits_{k=0}^\infty \frac{\left(-1\right)^{k+1}x^k}{(k+2)!3^{2k}}

    Copy my formula putting the mouse's pointer on it, and don't forget to open ALWAYS with [tex] and close with [/tex]

    Now, what n=th term are you talking about? the series' general term or the sequence of partial sums of the series?

    Tonio
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