Difference between revisions of "Mock AIME 3 Pre 2005 Problems/Problem 6"
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Notice that <math>\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)</math>. Thus, we have | Notice that <math>\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)</math>. Thus, we have | ||
− | < | + | <math></math>\begin{align*} |
\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}} \\ | \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) \ | + | &= \frac{1}{\sqrt{2}}\sum_{n = 1}^{9800} \left(\sqrt{n+1}-\sqrt{n-1}\right) \$<math> |
− | This is a [[telescope|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}< | + | This is a [[telescope|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}$. |
− | ==See | + | ==See Also== |
{{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}} | {{Mock AIME box|year=Pre 2005|n=3|num-b=5|num-a=7|source=19349}} | ||
[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] |
Revision as of 10:23, 4 April 2012
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
$$ (Error compiling LaTeX. Unknown error_msg)\begin{align*}
\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) \$\frac{1}{\sqrt{2}}\left(\sqrt{9801}+\sqrt{9800}-\sqrt{1}-\sqrt{0}\right) = 70 + 49\sqrt{2}
p+q+r=\boxed{121}$.
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 |