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aceace
Group Title
If T_{n} = \frac{ n1 }{ n }
, prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }
 one year ago
 one year ago
aceace Group Title
If T_{n} = \frac{ n1 }{ n } , prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }
 one year ago
 one year ago

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aceace Group TitleBest ResponseYou've already chosen the best response.0
If T_{n} = \frac{ n1 }{ n }, prove T_{n+1}  T _{n1} = \frac{ 2 }{ n^2 1 }
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
First, 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?
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Right, and \(T_{n1}\)?
 one year ago

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

aceace Group TitleBest ResponseYou've already chosen the best response.0
yeh i did that and got to the answer 2/(1n) which is wrong
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
Alright, 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?
 one year ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
yeh i got that step except i cant get the final answer
 one year ago

aceace Group TitleBest ResponseYou've already chosen the best response.0
its ok i got the answer... i had a silly mistake
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.1
We all make those from time to time.
 one year ago
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