<s>[INST] Determine the sum $T=a_0+a_1+a_2+...+a_{2012}$ Let ${a_n}$, $n \ge 0$ be a sequence of positive real numbers, given by
$a_0=1$ and
$a_m<a_n$ for all $m,n \in \mathbb{N}, m<n$ with 
$a_n=\sqrt{a_{n+1}a_{n-1}}+1$ and $4\sqrt{a_n}=a_{n+1}-a_{n-1}$ for all $n \in \mathbb{N}, n\neq 0$.
Help me, determining the sum $T=a_0+a_1+a_2+...+a_{2012}$.
 [/INST] $(a_{n+1}+a_{n-1})^2=(a_{n+1}-a_{n-1})^2+4a_{n+1}.a_{n-1}$
$=16a_n+4(a_n-1)^2=4(a_n+1)^2$
=>$a_{n+1}+a_{n-1}=2(a_n+1)$  as $a_n$ is increases with n and $a_0=1$
Or, $a_{n+1}-2 a_n +a_{n-1} -2 =0$
Putting n=m+1 and m,
$a_{m+2}-2 a_{m+1} +a_{m} -2 =0$  and 
$a_{m+1}-2 a_m +a_{m-1} -2 =0$
Subtracting $a_{m+2}-3a_{m+1}+3a_m-a_{m-1}=0$
If $a_m=b^m$, then $(b-1)^3=0$
So, $a_m=(Am^2+Bm+C)1^m=(Am^2+Bm+C)$
1=$a_0=C$
Following avatar, 4 = $a_1=A+B+C$  and 9 = $a_2=4A+2B+C =>(A,B,C)=(1,2,1)$
=>$a_m=(m^2+2m+1)=(m+1)^2$
Now, it should not be too tough.
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