Finding $S_{2013}^2$ from a Given Sequence

[Albert1]

a sequence ${a_n>0}$ for all n

given :$S_n - \dfrac{1}{a_n}=a_n-S_n$

find :$S_{2013}^2$

where :$S_n=a_1+a_2+-------+a_n$

 

[lfdahl]

Spoiler

One can easily find the first terms in the series:
\[a_1=\sqrt{1}-\sqrt{0} \\ a_2 = \sqrt{2}-\sqrt{1} \\ a_3 = \sqrt{3}-\sqrt{2}\]
Suppose this system holds for the nth level (n > 3): $a_n = \sqrt{n}-\sqrt{n-1}$ and let´s find $a_{n+1}$:
\[2S_{n+1} = a_{n+1}+\frac{1}{a_{n+1}} \\\\ 2(\sqrt{n}+a_{n+1}) = a_{n+1}+\frac{1}{a_{n+1}} \\\\ a_{n+1}^2 + 2\sqrt{n}\cdot a_{n+1}-1 = 0 \\\\ a_{n+1} = \frac{1}{2}\left ( -2\sqrt{n}\; \pm \sqrt{4n+4} \right ) \\\\ a_{n+1} = \sqrt{n+1}-\sqrt{n}\]
By the principle of induction $a_n=\sqrt{n}-\sqrt{n-1}$ and $S_n=\sqrt{n}$ for $n = 1,2,3,..$
Thus $S_{2013}^2 = (\sqrt{2013})^2 = 2013$

 

[Albert1]

Spoiler

One can easily find the first terms in the series:
\[a_1=\sqrt{1}-\sqrt{0} \\ a_2 = \sqrt{2}-\sqrt{1} \\ a_3 = \sqrt{3}-\sqrt{2}\]
Suppose this system holds for the nth level (n > 3): $a_n = \sqrt{n}-\sqrt{n-1}$ and let´s find $a_{n+1}$:
\[2S_{n+1} = a_{n+1}+\frac{1}{a_{n+1}} \\\\ 2(\sqrt{n}+a_{n+1}) = a_{n+1}+\frac{1}{a_{n+1}} \\\\ a_{n+1}^2 + 2\sqrt{n}\cdot a_{n+1}-1 = 0 \\\\ a_{n+1} = \frac{1}{2}\left ( -2\sqrt{n}\; \pm \sqrt{4n+4} \right ) \\\\ a_{n+1} = \sqrt{n+1}-\sqrt{n}\]
By the principle of induction $a_n=\sqrt{n}-\sqrt{n-1}$ and $S_n=\sqrt{n}$ for $n = 1,2,3,..$
Thus $S_{2013}^2 = (\sqrt{2013})^2 = 2013$

very good solution :)