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Thread: Sequences

  1. #1
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    Sequences

    Gven Un=6n-3-a, where a is constant,to find the recurrence relation of the form Un+1=f(Un),what is my first step?any suggestion?i dont even know how to start





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  2. #2
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    Re: Sequences

    Try $U_{n+1}-U_n=6(n+1)-3-a-(6n-3-a)$

    Add $U_n $ to both sides and simplify.
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  3. #3
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    Re: Sequences

    Quote Originally Posted by goldsomecriss View Post
    Gven Un=6n-3-a, where a is constant,to find the recurrence relation of the form Un+1=f(Un),what is my first step?any suggestion?i dont even know how to start
    As I read this question, $U_n=6n-3-a.~\forall n$.
    $ \begin{align*}U_1&=3-a\\U_2&=9-a\\U_3&=15-a\end{align*}$.

    So$ \begin{align*} U_{n+1} &= 6(n+1)-3-a \\ &=(6n-3-a)+6\\&=U_n+6\end{align*}$
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    Re: Sequences

    Quote Originally Posted by Plato View Post
    As I read this question, $U_n=6n-3-a.~\forall n$.
    $ \begin{align*}U_1&=3-a\\U_2&=9-a\\U_3&=15-a\end{align*}$.

    So$ \begin{align*} U_{n+1} &= 6(n+1)-3-a \\ &=(6n-3-a)+6\\&=U_n+6\end{align*}$
    thanks

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