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aceace

  • 2 years ago

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

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

  2. KingGeorge
    • 2 years ago
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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?

  3. aceace
    • 2 years ago
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    n/n+1

  4. KingGeorge
    • 2 years ago
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    Right, and \(T_{n-1}\)?

  5. aceace
    • 2 years ago
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    n-2/n-1

  6. KingGeorge
    • 2 years ago
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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.

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

  8. KingGeorge
    • 2 years ago
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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?

  9. aceace
    • 2 years ago
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    just a sec

  10. aceace
    • 2 years ago
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    yeh i got that step except i cant get the final answer

  11. aceace
    • 2 years ago
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    its ok i got the answer... i had a silly mistake

  12. KingGeorge
    • 2 years ago
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    We all make those from time to time.

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