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aceace Group Title

If T_{n} = \frac{ n-1 }{ n } , prove T_{n+1} - T _{n-1} = \frac{ 2 }{ n^2 -1 }

  • one year ago
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  1. aceace Group Title
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    If T_{n} = \frac{ n-1 }{ n }, prove T_{n+1} - T _{n-1} = \frac{ 2 }{ n^2 -1 }

    • one year ago
  2. KingGeorge Group Title
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    First, you want to write out what \(T_{n+1}\) and \(T_{n-1}\) are using the fact that \(\displaystyle T_n=\frac{n-1}{n}\). Can you tell me what \(T_{n+1}\) is?

    • one year ago
  3. aceace Group Title
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    n/n+1

    • one year ago
  4. KingGeorge Group Title
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    Right, and \(T_{n-1}\)?

    • one year ago
  5. aceace Group Title
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    n-2/n-1

    • one year ago
  6. KingGeorge Group Title
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    Bingo. So you have \[T_{n+1}-T_{n-1}=\frac{n}{n+1}-\frac{n-2}{n-1}\]Find a common denominator, and simplify.

    • one year ago
  7. aceace Group Title
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    yeh i did that and got to the answer 2/(1-n) which is wrong

    • one year ago
  8. KingGeorge Group Title
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    Alright, to get a common denominator, you do the following, and simplify. \[\left(\frac{n}{n+1}\cdot\frac{n-1}{n-1}\right)-\left(\frac{n-2}{n-1}\cdot\frac{n+1}{n+1}\right)\] Does this help?

    • one year ago
  9. aceace Group Title
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    just a sec

    • one year ago
  10. aceace Group Title
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    yeh i got that step except i cant get the final answer

    • one year ago
  11. aceace Group Title
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    its ok i got the answer... i had a silly mistake

    • one year ago
  12. KingGeorge Group Title
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    We all make those from time to time.

    • one year ago
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