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Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"

(→‎Solution 2)
(→‎Solution 2)
 
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~ Nafer
 
~ Nafer
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Let <math>P(x)=x^4-3x^3-ax^2+bx-12</math> such that
 
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<cmath>P(x)=(x-1)(x-2)(x-3)(x-m)+k</cmath>
 
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for some <math>m</math> and <math>f(1)=f(2)=f(3)=k</math>. Expanding gets
 
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<cmath>P(x)=x^4-(m+6)x^3+(6m+11)x^2-(11m+6)x+6m+k</cmath>
 
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Using the corresponding coefficient of <math>x^3</math> , we have
 
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<cmath>m+6=3\Longrightarrow m=-3</cmath>
 
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Thus
 
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<cmath>P(x)=x^4-3x^3-7x^2+27^x-12</cmath>
 
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By long division,  <math>f(x)=48</math> (<math>\boxed{C})</math>.
 
  
 
==See Also==
 
==See Also==

Latest revision as of 14:48, 29 November 2019

Problem

Let $S$ denote the value of the sum

\[\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\]

$S$ can be expressed as $p + q \sqrt{r}$, where $p, q,$ and $r$ are positive integers and $r$ is not divisible by the square of any prime. Determine $p + q + r$.

Solution

Notice that $\sqrt{n + \sqrt{n^2 - 1}} = \frac{1}{\sqrt{2}}\sqrt{2n + 2\sqrt{(n+1)(n-1)}} = \frac{1}{\sqrt{2}}\left(\sqrt{n+1}+\sqrt{n-1}\right)$. Thus, we have

\[\sum_{n = 1}^{9800} \frac{1}{\sqrt{n + \sqrt{n^2 - 1}}}\] \[= \sqrt{2}\sum_{n = 1}^{9800} \frac{1}{\sqrt{n+1}+\sqrt{n-1}}\] \[= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right)\]

This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with $\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}$, and $p+q+r=\boxed{121}$.


Solution 2

Simplifying the expression yields \begin{align*} S &= \sum_{n=1}^{9800}\frac{1}{\sqrt{n+\sqrt{n^2-1}}} \\ &= \sum_{n=1}^{9800}\frac{\sqrt{n+\sqrt{n^2-1}}}{n+\sqrt{n^2-1}} \\ &= \sum_{n=1}^{9800}(\frac{\sqrt{n+\sqrt{n^2-1}}}{n+\sqrt{n^2-1}})\cdot(\frac{n-\sqrt{n^2-1}}{n-\sqrt{n^2-1}}) \\ &= \sum_{n=1}^{9800}(\sqrt{n+\sqrt{n^2-1}})\cdot(\sqrt{n+\sqrt{n^2-1}})^2 \\ &= \sum_{n=1}^{9800}\sqrt{n-\sqrt{n^2-1}} \\ \end{align*} Now we can assume that \[\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}\] \[\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}\] for some $a$, $b$, $c$.

Squaring the first equation yields \[n+\sqrt{n^2-1}=a^2+b^2c+2ab\sqrt{c}\] which gives the system of equations \[n=a^2+b^2c\] \[\sqrt{n^2-1}=2ab\sqrt{c}\] calling them equations $A$ and $B$, respectively.

Also we have \[\frac{1}{\sqrt{n+\sqrt{n^2-1}}}=\sqrt{n-\sqrt{n^2-1}}\] \[\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}\] \[a^2-b^2c=1\] which obtains equation $C$.

Adding equations $A$ and $C$ yields \[2a^2=n+1\] \[a=\sqrt{\frac{n+1}{2}}\] Squaring equation $B$ and substituting yields \[4a^2b^2c=n^2-1\] \[2\cdot(n+1)\cdot b^2c=n^2-1\] \[b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}\] \[b\sqrt{c}=\sqrt{\frac{n-1}{2}}\]

Thus we obtain the telescoping series \begin{align*} S &= \sum_{n=1}^{9800}a-b\sqrt{c} \\ &= \sum_{n=1}^{9800}\sqrt{\frac{n+1}{2}}-\sqrt{\frac{n-1}{2}} \\ \end{align*}

Simplifying the sum we are left with \begin{align*} S &= -\sqrt{\frac{1}{2}}+\sqrt{\frac{9800}{2}}+\sqrt{\frac{9801}{2}} \\ &= -\frac{\sqrt{2}}{2}+\frac{99\sqrt{2}}{2}+70 \\ &= 70+49\sqrt{2} \\ \end{align*}

Thus $p+q+r=70+49+2=\boxed{121}$.


~ Nafer

See Also

Mock AIME 3 Pre 2005 (Problems, Source)
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Problem 5 		Followed by
Problem 7
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