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

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The discussion focuses on simplifying the expression for the sequence \( a_n \) using surds and rationalizing the denominator. The final form of \( a_n \) is derived as \( -\frac{1}{2}(\sqrt{n+1}+\sqrt{n-1}-2\sqrt{n}) \), which reveals it as a telescoping series. To compute \( S_{2012} \), the participants derive the expression \( S_{2012} = -\frac{1}{2}(\sqrt{2013}-\sqrt{1})+\frac{1}{2}(\sqrt{2012}-0) \). A correction is noted regarding a typo in the final step, where a \( \frac{1}{2} \) was omitted. The discussion emphasizes the importance of accuracy in mathematical expressions.
Albert1
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$ 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}) $
$S_n=a_1+a_2+a_3+-------+a_n$
$find:\,\, S_{2012}$
 
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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:

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

$$S_{2012}=\frac{1}{2}-\frac{1}{2}(\sqrt{2013}-\sqrt{2012})$$
 
Last edited:
anemone said:
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
 
Albert said:
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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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