<s>[INST] How can I find general formula of this sequence? Let be the sequence $(a_n)$ so that $a_1 = 1$,  $a_{n+1}=\dfrac{20}{3 a_n+4}$, $\forall n \geqslant 1.$ I can prove that the sequence is an increasing sequence and convert to 2. But, I can find the general formula of this sequence. How can I find it?
 [/INST] $$A_{n+1}=\frac{20}{3A_n+4} \implies A_{n+1}~(3A_n +4)= 20.~~~(1)$$
Let $$(3A_n+4)=\frac{B_n}{B_{n-1}} \implies A_n=\frac{1}{3} \left(\frac{B_n}{B_{n-1}}-4\right)~~~(2)$$
Then (1) becomes $$B_{n+1}-60 B_{n-1}-4 B_n =0~~~(3)$$
Now let us put $B_n=t^n$ in (3) to get 
$$t^2-4t-60=0 \implies t=10,-6.$$
So the solution of (3) is $$B_n=P (10)^n +Q(-6)^n ~~~(4)$$
Inserting (4) in (2) we get $$A_n=\frac{1}{3} \left(\frac{6 R (10)^{n-1}-10 (-6)^{n-1}}{R (10)^{n-1}+ (-6)^{n-1}}\right), R=\frac{P}{Q}.$$
Using $A_1=1$ gives $$R=\frac{13}{3}.$$
Finally, we get $$A_n= \left( \frac{26~ (5)^{n-1}-10 ~ (-3)^{n-1}}{13~ (5)^{n-1} +3 ~(-3)^{n-1}} \right).$$
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