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 $T=a_0+a_1+a_2+...+a_{2012}$.
 A: Take $n=1$ in both the equations, you will get, $a_1=\sqrt {a_2}+1$ and $4\sqrt {a_1}=a_2-1$ (here i substituted $a_0=1$), solving these equations gives $a_1=4=2^2$ and $a_2=9=3^2$. Now you can calculate $a_3$ by putting $n=2$ in any one of the given equations(the equations are consistent(you can check by using both to obtain answer)) which gives you $a_3=16=4^2$. Following this manner, you will get further terms as $25,36,49\cdots {2013}^2$. Therefore sum is $1^2+2^2+3^2+\cdots +{2013}^2=2721031819$
A: 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 $n=2012$.
A: $(a_{n+1}+a_{n-1})^2=(a_{n+1}-a_{n-1})^2+4a_{n+1}.a_{n-1}$
$=16a_n+4(a_n-1)^2=4(a_n+1)^2$
=>$a_{n+1}+a_{n-1}=2(a_n+1)$  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.
