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Math Help - Sequences and series question.

  1. #1
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    Sequences and series question.

    Hi guys, rhe following is a question that I busted my brains over, but could not get, so help would be greatly appreciated.

    And I don't know how to properly use latex, so please bear with me.

    Q) A sequence satisfies T_{n}\=\frac{\1}{2}(T_{n-1}\+\T_{n+1}) with T_{1} = 3 and T_{2} = 2.

    Find T_{3} and T_{4}

    Also, T1=2 and T2=7

    Question without latex: A sequence satisfies [term 'n' = 0.5*[term 'n-1' + term 'n+1']. Find term's 3 and 4.

    Sorry I have absolutely no idea how to do this, one, also my latex skills are pretty poor, so ANY HELP WOULD BE GREATLY APPRECIATED.
    Last edited by mr fantastic; October 11th 2010 at 04:12 PM. Reason: Re-titled.
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  2. #2
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    T_{n}=\frac{1}{2}(T_{n-1}+T_{n+1}) with T_{1} = 3 and T_{2} = 2 .

    When n = 2, the problem becomes

    T_{2}=\frac{1}{2}(T_{1}+T_{3})

    Substitute the known values and find T3.
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  3. #3
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    End the LaTex with /math, not \math.
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  4. #4
    MHF Contributor chisigma's Avatar
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    The difference equation written in 'standard form' is...

    t_{n} - 2\ t_{n-1} + t_{n-2}=0 , t_{0}= 3\ ,\ t_{1}=2 (1)

    ... and its general solution depends from the root of algebraic equation...

    x^{2} - 2\ x +1=0 (2)

    ... i.e. x=1 with multeplicity 2. The general solution is then...

    t_{n}= c_{1} + n\ c_{2} (3)

    ... that with the 'initial conditions' becomes...

    t_{n} = 3 - n (4)

    Kind regards

    \chi \sigma
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  5. #5
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    Hello, SeaNanners


    \text{A sequence satisfies: }\;T_{n}\:=\:\frac{1}{2}(T_{n-1}+T_{n+1})\;\text{ with }\,T_{1} = 3\,\text{ and }\,T_{2} = 2

    \text{Find }\,T_{3}\,\text{ and }\,T_{4}

    Each term is the average of the term before and the term after.


    We have: . \begin{array}{|c|c|c|c|c}<br />
T_1 & T_2 & T_3 & T_4 & \hdots \\ \hline<br />
3 & 2 & ? & \end{array}

    \,T_2 is the average of \,T_1 and T_3.

    . . 2 \:=\:\dfrac{3+T_3}{2} \quad\Rightarrow\quad \boxed{T_3 \:=\:1}


    We have: . \begin{array}{|c|c|c|c|c}<br />
T_1 & T_2 & T_3 & T_4 & \hdots \\ \hline<br />
3 & 2 & 1 & ? \end{array}

    T_3 is the average of T_2 and T_4.

    . . 1 \:=\:\dfrac{2+T_4}{2} \quad\Rightarrow\quad \boxed{T_4 \:=\:0}
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