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I tried a lot of different methods. Not able to make out the series. Could anyone help me i this regard?

$ T(n) = \frac{(n+1)}{n}T(n-1) + c\frac{(2n-1)}{n} , T(1) = 0 $

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I have edited Just tell if the T(n-1) is in numerator or denominator –  sammath Feb 22 at 7:39
    
If you write the first terms, you should notice that, for any $n$, $T(n)$ is proportional to $c$. –  Claude Leibovici Feb 22 at 8:28
    
Have a look at my answer to math.stackexchange.com/questions/684946/…, this is more-or-less the same problem. –  Jack D'Aurizio Feb 22 at 10:03

1 Answer 1

$$\frac{T(n)}{n+1}=\frac{T(n-1)}{n}+c\frac{2n-1}{n(n+1)}=\frac{T(n-1)}{n}+\frac{3c}{n+1}-\frac{c}n$$

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