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anonymous
 3 years ago
If T_{n} = \frac{ n1 }{ n }
, prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }
anonymous
 3 years ago
If T_{n} = \frac{ n1 }{ n } , prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If T_{n} = \frac{ n1 }{ n }, prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1First, you want to write out what \(T_{n+1}\) and \(T_{n1}\) are using the fact that \(\displaystyle T_n=\frac{n1}{n}\). Can you tell me what \(T_{n+1}\) is?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Right, and \(T_{n1}\)?

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Bingo. So you have \[T_{n+1}T_{n1}=\frac{n}{n+1}\frac{n2}{n1}\]Find a common denominator, and simplify.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeh i did that and got to the answer 2/(1n) which is wrong

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1Alright, to get a common denominator, you do the following, and simplify. \[\left(\frac{n}{n+1}\cdot\frac{n1}{n1}\right)\left(\frac{n2}{n1}\cdot\frac{n+1}{n+1}\right)\] Does this help?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeh i got that step except i cant get the final answer

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0its ok i got the answer... i had a silly mistake

KingGeorge
 3 years ago
Best ResponseYou've already chosen the best response.1We all make those from time to time.
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