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 $T=a_0+a_1+a_2+...+a_{2012}$.

From the conditions given, we have
$$
(a_n-1)^2=a_{n+1}a_{n-1}\tag{1}
$$
and
$$
16a_n=(a_{n+1}-a_{n-1})^2\tag{2}
$$
Adding $4$ times $(1)$ to $(2)$ yields
$$
4(a_n+1)^2=(a_{n+1}+a_{n-2})^2\tag{3}
$$
Taking the square root of each side and subtracting $2a_n$ from both sides, we get
$$
a_{n+1}-2a_n+a_{n-1}=2\tag{4}
$$
Thus, because the second difference of $a_n$ is $2$, $a_n=n^2+bn+c$. Since $a_0=1$, we get that $c=1$. Plugging $n^2+bn+1$ into either $(1)$ or $(2)$ gives $b^2=4$.  Since $a_1>a_0$, we must have $b=2$. That is,
$$
a_n=(n+1)^2\tag{5}
$$
To sum consecutive squares, use
$$
\begin{align}
\sum_{k=0}^{n}(k+1)^2
&=\frac{(2n+3)(n+2)(n+1)}{6}\\
&=2721031819\tag{6}
\end{align}
$$
for $n=2012$.
