2018 MPFG Problem 19: Difference between revisions
| (3 intermediate revisions by the same user not shown) | |||
| Line 7: | Line 7: | ||
==Solution 1== | ==Solution 1== | ||
We can think of this problem through integration perspectives. Observe that < | We can think of this problem through integration perspectives. Observe that <imath>S_n</imath> looks very similar to a Riemann sum. | ||
<cmath> | <cmath>S_n = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{3}}+ ... + \frac{1}{\sqrt{9801}}</cmath> | ||
We first applicate the right Riemann sum of < | We first applicate the right Riemann sum of <imath>y=\frac{1}{\sqrt{x}}</imath> | ||
[[File:Right_rie.jpg|750px|center]] | [[File:Right_rie.jpg|750px|center]] | ||
<cmath>2S_n > \ | <cmath>2S_n > \int_{1}^{9803} \frac{1}{\sqrt{x}} \,dx</cmath> | ||
<cmath>2S_n > \left. (2x^{\frac{1}{2}})\right|_{1}^{9803}</cmath> | <cmath>2S_n > \left. (2x^{\frac{1}{2}})\right|_{1}^{9803}</cmath> | ||
| Line 23: | Line 23: | ||
<cmath>S_n > \sqrt{9803}-1</cmath> | <cmath>S_n > \sqrt{9803}-1</cmath> | ||
Then applicate the left Riemann sum of < | Then applicate the left Riemann sum of <imath>y=\frac{1}{\sqrt{x}}</imath> | ||
[[File:Left_rie.jpg|750px|center]] | [[File:Left_rie.jpg|750px|center]] | ||
<cmath>2S_n-1 < \ | <cmath>2S_n-1 < \int_{1}^{9801} \frac{1}{\sqrt{x}} \,dx</cmath> | ||
<cmath>2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}</cmath> | <cmath>2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}</cmath> | ||
| Line 33: | Line 33: | ||
<cmath>S_n-\frac{1}{2} > \sqrt{9801}-1</cmath> | <cmath>S_n-\frac{1}{2} > \sqrt{9801}-1</cmath> | ||
<cmath>S_n < \sqrt{9801}-1 | <cmath>S_n < \sqrt{9801}-\frac{1}{2}</cmath> | ||
We conclude that: | We conclude that: | ||
< | <imath>\sqrt{9803}-1 < S_n < \sqrt{9801}-\frac{1}{2}</imath> | ||
< | <imath>\lfloor S_n \rfloor = \boxed{98}</imath> | ||
~cassphe | ~cassphe | ||
Latest revision as of 09:29, 7 November 2025
Problem 19
Consider the sum
![\[S_n = \sum_{k=1}^{n}\frac{1}{\sqrt{2k-1}}\]](http://latex.artofproblemsolving.com/2/4/0/240b62e32ea68da7b45be64107ac9da7c52d1b8e.png)
Determine 
. Recall that if 
is a real number, then 
(the floor of x) is the greatest integer that is less than or equal to .
Solution 1
We can think of this problem through integration perspectives. Observe that 
looks very similar to a Riemann sum.
![\[S_n = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{3}}+ ... + \frac{1}{\sqrt{9801}}\]](http://latex.artofproblemsolving.com/a/9/a/a9a0e004d369f1c342d1f3920467e6a74fcb875d.png)
We first applicate the right Riemann sum of 
![\[2S_n > \int_{1}^{9803} \frac{1}{\sqrt{x}} \,dx\]](http://latex.artofproblemsolving.com/9/5/7/957a83ba7f7f19e7d568c0c3f7a81c3e770a8de9.png)
![\[2S_n > \left. (2x^{\frac{1}{2}})\right|_{1}^{9803}\]](http://latex.artofproblemsolving.com/3/f/2/3f239abd737ba3da55d27fd1af4afc2ea033ff71.png)
![\[2S_n > 2(\sqrt{9803}-1)\]](http://latex.artofproblemsolving.com/d/5/b/d5b9cabd1b9220136c997c436b0f9c1e9fcf8ff0.png)
![\[S_n > \sqrt{9803}-1\]](http://latex.artofproblemsolving.com/c/c/8/cc8a19de16c79457398d7200f4e7d3372111a206.png)
Then applicate the left Riemann sum of
![\[2S_n-1 < \int_{1}^{9801} \frac{1}{\sqrt{x}} \,dx\]](http://latex.artofproblemsolving.com/9/4/5/94579f10b7da9091f05dd8ef18c37a3f654589a8.png)
![\[2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}\]](http://latex.artofproblemsolving.com/4/1/3/41370ace1d9c0be85df031a0c95411d63b92bc67.png)
![\[S_n-\frac{1}{2} > \sqrt{9801}-1\]](http://latex.artofproblemsolving.com/8/1/c/81c52b56cae20dac843c916a2ceaf4261c215970.png)
![\[S_n < \sqrt{9801}-\frac{1}{2}\]](http://latex.artofproblemsolving.com/f/9/9/f99c7d4799faa9abb3e3ab9cde7f31fc0f74d89b.png)
We conclude that:
~cassphe
