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.
 
I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...

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