How Do I Simplify Fractions Using Surds in the Summation Formula?

[Albert1]

$ a_n=(\dfrac{1}{\sqrt n+\sqrt {n-1}})\times(\dfrac{1}{\sqrt {n+1}+\sqrt {n-1}})\times(\dfrac{1}{\sqrt {n+1}+\sqrt n}) $
$S_n=a_1+a_2+a_3+-------+a_n$
$find:\,\, S_{2012}$

 

[anemone]

Rationalizing the denominator of the expression $$a_n$$ we get:

$ a_n=\left(\dfrac{1}{\sqrt n+\sqrt {n-1}}\cdot \dfrac {\sqrt n-\sqrt {n-1}}{\sqrt n-\sqrt {n-1}}\right)\times\left(\dfrac{1}{\sqrt {n+1}+\sqrt {n-1}}\cdot \dfrac {\sqrt {n+1}-\sqrt {n-1}}{\sqrt {n+1}-\sqrt {n-1}}\right)\times\left(\dfrac{1}{\sqrt {n+1}+\sqrt {n}}\cdot \dfrac {\sqrt {n+1}-\sqrt {n}}{\sqrt {n+1}-\sqrt {n}}\right) $

$ a_n=\left(\dfrac {\sqrt {n}-\sqrt {n-1}}{1}\right)\times \left(\dfrac {\sqrt {n}-\sqrt {n+1}}{-1}\right) \times \left(\dfrac {\sqrt {n+1}-\sqrt {n-1}}{2}\right)$$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}\right)$

$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}-\sqrt{n}\right)+\dfrac{1}{2}\left(\sqrt{n}-\sqrt{n-1}\right)$

This is clearly a telescoping series and to compute $$S_{2012}$$, we get:

$$S_{2012}=-\frac{1}{2}(\sqrt {2013}-\sqrt {1})+\frac{1}{2}(\sqrt{2012}-0)$$

$$S_{2012}=\frac{1}{2}-\frac{1}{2}(\sqrt{2013}-\sqrt{2012})$$

 

[Albert1]

Rationalizing the denominator of the expression $$a_n$$ we get:$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}\right)$

$ a_n=-\dfrac{1}{2}\left(\sqrt{n+1}-\sqrt{n}\right)+\dfrac{1}{2}\left(\sqrt{n}-\sqrt{n-1}\right)$

This is clearly a telescoping series and to compute $$S_{2012}$$, we get:

$$S_{2012}=-\frac{1}{2}(\sqrt {2013}-\sqrt {1})+\frac{1}{2}(\sqrt{2012}-0)$$

$$S_{2012}=\frac{1}{2}-(\sqrt{2013}-\sqrt{2012})-------(last \,\, step)$$

your last step is incorrect ,a typo happens

 

[anemone]

your last step is incorrect ,a typo happens

Yep, you're right Albert...I left off $$\frac{1}{2}$$ in front of the surds, I'm sorry and I will fix my first post so that I get the correct answer to this problem.