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

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2012 is a very interesting number. – peoplepower Jul 26 '12 at 12:00
What did you try? – Did Jul 26 '12 at 12:03

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 $n=2012$.

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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 $1^2+2^2+3^2+\cdots +{2013}^2=2721031819$

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$\displaystyle \sum_{k=1}^{n} k^2= \frac16 n(n+1)(2n+1)$, but can you please explain in more detail how you got there? – draks ... Jul 26 '12 at 19:31
I calculated first few terms as written in my answer up-to $a_2$, where i observe the pattern $1^2,2^2,3^2,..$ for $n=0,1,2,..$. so, i thought general term $a_n$ might be $(n+1)^2$. I put this substitution in your equations, and they satisfied it. – Aang Jul 27 '12 at 5:56

$(a_{n+1}+a_{n-1})^2=(a_{n+1}-a_{n-1})^2+4a_{n+1}.a_{n-1}$

$=16a_n+4(a_n-1)^2=4(a_n+1)^2$

=>$a_{n+1}+a_{n-1}=2(a_n+1)$ 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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