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If T_{n} = \frac{ n-1 }{ n } , prove T_{n+1} - T _{n-1} = \frac{ 2 }{ n^2 -1 }
If T_{n} = \frac{ n-1 }{ n }, prove T_{n+1} - T _{n-1} = \frac{ 2 }{ n^2 -1 }
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?
Right, and \(T_{n-1}\)?
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.
yeh i did that and got to the answer 2/(1-n) which is wrong
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?
yeh i got that step except i cant get the final answer
its ok i got the answer... i had a silly mistake
We all make those from time to time.