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Math Help - Recurrence/APGP/Sigma Notation

  1. #1
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    Recurrence/APGP/Sigma Notation

    1) Un is defined by recurrence relation:
    u1=2 and u(n)=u(n-1)+1-1/(n(n-1)) for n>=2

    By using method of diff, find u(n) (Ans: (n^2)/(n+1))

    Some quick help wud be great! thanks
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  2. #2
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    Hello, zeromeyzl!

    Could you restate the problem?
    As given, it doesn't compute . . .


    u(n) is defined by recurrence relation:
    . . u(1)\:=\:2,\;\;u(n)\:=\:u(n-1)+1-\frac{1}{n(n-1)}\;\;\text{ for }n \geq 2

    By using method of diff, find u(n)

    Answer: . \frac{n^2}{n+1} . . . . not correct

    If I read everything correctly, we have:

    . . u(1) \;=\;2

    . . u(2) \;=\;2 + 1 - \frac{1}{2\cdot1} \;=\;3 - \frac{1}{2} \;=\;\frac{5}{2}

    . . u(3) \;=\;\frac{5}{2} + 1 - \frac{1}{3\cdot2} \;=\;\frac{7}{2} - \frac{1}{6} \;=\;\frac{20}{6} \;=\;\frac{10}{3}<br />

    . . u(4) \;=\;\frac{10}{3} + 1 - \frac{1}{4\cdot3} \;=\;\frac{13}{3} - \frac{1}{12} \;=\;\frac{51}{12} \;=\;\frac{17}{4}


    The general term for this sequence is: . u(n) \;=\;\frac{n^2+1}{n}

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