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$a_{n+1} = \frac{1+a_{n}}{3-a_{n}},n\in \mathbb{Z}^{+}$, and $a_{1} = 0$, so how to get the general expression of $a_{n} = ?$

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Hint: have you tried looking at the first few $n$'s? You should see a pattern emerge pretty quickly.

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As suggested by Robert Israel, let us look at the first terms so we might see a pattern: we have $a_1=0$, $a_2=\frac {1}{3}$, $a_3=\frac {2}{4}$, $a_4=\frac {3}{5}$. All of this suggest $$a_n=\frac{n-1}{n+1}$$

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Evaluating the first few terms of the sequence: $$a_1=0, a_2=\frac{1}{3}, a_3=\frac{2}{4}, a_4=\frac{3}{5}, a_5=\frac{4}{6}, a_6=\frac{5}{7} \dots$$

So the guess is $a_n=\frac{n-1}{n+1}$. Now let's prove it using induction

Clearly it's true for $n=1$. Assuming the statement is true for $n=k$, i.e. $a_k=\frac{k-1}{k+1}$ $$a_{k+1}=\frac{1+a_k}{3-a_k}=\frac{(k+1)-1}{(k+1)+1}$$

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  • $\begingroup$ I think we disagree. Check my answer, please. Cheers. $\endgroup$ – Claude Leibovici Apr 13 '14 at 9:57
  • $\begingroup$ @ClaudeLeibovici thanks a lot for pointing the error. please note the correction and sorry for not checking. $\endgroup$ – Sandeep Thilakan Apr 13 '14 at 12:23

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