Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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This is a [[telescoping series]]; note that when we expand the summation, all of the intermediary terms cancel, leaving us with | This is a [[telescoping series]]; note that when we expand the summation, all of the intermediary terms cancel, leaving us with | ||
<math>\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}</math>, and <math>p+q+r=\boxed{121}</math>. | <math>\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}</math>, and <math>p+q+r=\boxed{121}</math>. | ||
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+ | |||
== Solution 2 == | == Solution 2 == | ||
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&= \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}} \\ | ||
&= \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}(\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}})\cdot(\sqrt{n+\sqrt{n^2-1}})^2 \\ |
+ | &= \sum_{n=1}^{9800}\sqrt{n-\sqrt{n^2-1}} \\ | ||
+ | \end{align*}</cmath> | ||
+ | Now we can assume that | ||
+ | <cmath>\sqrt{n+\sqrt{n^2-1}}=a+b\sqrt{c}</cmath> | ||
+ | <cmath>\sqrt{n-\sqrt{n^2-1}}=a-b\sqrt{c}</cmath> | ||
+ | for some <math>a</math>, <math>b</math>, <math>c</math>. | ||
+ | |||
+ | Squaring the first equation yields | ||
+ | <cmath>n+\sqrt{n^2-1}=a^2+b^2c+2ab\sqrt{c}</cmath> | ||
+ | which gives the system of equations | ||
+ | <cmath>n=a^2+b^2c</cmath> | ||
+ | <cmath>\sqrt{n^2-1}=2ab\sqrt{c}</cmath> | ||
+ | calling them equations <math>A</math> and <math>B</math>, respectively. | ||
+ | |||
+ | Also we have | ||
+ | <cmath>\frac{1}{\sqrt{n+\sqrt{n^2-1}}}=\sqrt{n-\sqrt{n^2-1}}</cmath> | ||
+ | <cmath>\frac{1}{a+b\sqrt{c}}=a-b\sqrt{c}</cmath> | ||
+ | <cmath>a^2-b^2c=1</cmath> | ||
+ | which obtains equation <math>C</math>. | ||
+ | |||
+ | Adding equations <math>A</math> and <math>C</math> yields | ||
+ | <cmath>2a^2=n+1</cmath> | ||
+ | <cmath>a=\sqrt{\frac{n+1}{2}}</cmath> | ||
+ | Squaring equation <math>B</math> and substituting yields | ||
+ | <cmath>4a^2b^2c=n^2-1</cmath> | ||
+ | <cmath>2\cdot(n+1)\cdot b^2c=n^2-1</cmath> | ||
+ | <cmath>b^2c=\frac{(n-1)(n+1)}{2\cdot(n+1)}</cmath> | ||
+ | <cmath>b\sqrt{c}=\sqrt{\frac{n-1}{2}}</cmath> | ||
+ | |||
+ | Thus we obtain the telescoping series | ||
+ | <cmath>\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*}</cmath> | ||
+ | |||
+ | Simplifying the sum we are left with | ||
+ | <cmath>\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*}</cmath> | \end{align*}</cmath> | ||
+ | |||
+ | Thus <math>p+q+r=70+49+2=\boxed{121}</math>. | ||
+ | |||
+ | |||
+ | ~ Nafer | ||
==See Also== | ==See Also== |
Latest revision as of 14:48, 29 November 2019
Contents
Problem
Let denote the value of the sum
can be expressed as
, where
and
are positive integers and
is not divisible by the square of any prime. Determine
.
Solution
Notice that . Thus, we have
This is a telescoping series; note that when we expand the summation, all of the intermediary terms cancel, leaving us with
, and
.
Solution 2
Simplifying the expression yields
Now we can assume that
for some
,
,
.
Squaring the first equation yields
which gives the system of equations
calling them equations
and
, respectively.
Also we have
which obtains equation
.
Adding equations and
yields
Squaring equation
and substituting yields
Thus we obtain the telescoping series
Simplifying the sum we are left with
Thus .
~ Nafer
See Also
Mock AIME 3 Pre 2005 (Problems, Source) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |