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2018 MPFG Problem 19: Difference between revisions

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==Solution 1==
==Solution 1==
We can think of this problem through integration perspectives. Observe that <math>S_n</math> looks very similar to a Riemann sum.
We can think of this problem through integration perspectives. Observe that <imath>S_n</imath> looks very similar to a Riemann sum.


<cmath>_n = 1/\sqrt{1}+1/\sqrt{3}+ ... + 1/\sqrt{9801}</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 <math>y=\frac{1}{\sqrt{x}}</math>
We first applicate the right Riemann sum of <imath>y=\frac{1}{\sqrt{x}}</imath>


[insert pic]
[[File:Right_rie.jpg|750px|center]]


<cmath>2S_n > \int{f_{1}^{9803}(\frac{1}{\sqrt{x}})}dx</cmath>
<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 <math>y=\frac{1}{\sqrt{x}}</math>
Then applicate the left Riemann sum of <imath>y=\frac{1}{\sqrt{x}}</imath>


[insert pic2]
[[File:Left_rie.jpg|750px|center]]


<cmath>2(S_n-1) < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx</cmath>
<cmath>2S_n-1 < \int_{1}^{9801} \frac{1}{\sqrt{x}} \,dx</cmath>


<cmath>2(S_n-1) < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}</cmath>
<cmath>2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}</cmath>


<cmath>S_n-1 > \sqrt{9801}-1</cmath>
<cmath>S_n-\frac{1}{2} > \sqrt{9801}-1</cmath>


<cmath>S_n < \sqrt{9801}</cmath>
<cmath>S_n < \sqrt{9801}-\frac{1}{2}</cmath>




We conclude that:
We conclude that:


<math>\sqrt{9803}-1 < S_n < \sqrt{9801}</math>
<imath>\sqrt{9803}-1 < S_n < \sqrt{9801}-\frac{1}{2}</imath>


<math>\lfloor S_n \rfloor = \boxed{98}</math>
<imath>\lfloor S_n \rfloor = \boxed{98}</imath>


~cassphe
~cassphe

Latest revision as of 09:29, 7 November 2025

Problem 19

Consider the sum

\[S_n = \sum_{k=1}^{n}\frac{1}{\sqrt{2k-1}}\]

Determine $\lfloor S_{4901} \rfloor$. Recall that if $x$ is a real number, then $\lfloor x \rfloor$ (the floor of x) is the greatest integer that is less than or equal to $x$.

Solution 1

We can think of this problem through integration perspectives. Observe that $S_n$ looks very similar to a Riemann sum.

\[S_n = \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{3}}+ ... + \frac{1}{\sqrt{9801}}\]

We first applicate the right Riemann sum of $y=\frac{1}{\sqrt{x}}$

\[2S_n > \int_{1}^{9803} \frac{1}{\sqrt{x}} \,dx\]

\[2S_n > \left. (2x^{\frac{1}{2}})\right|_{1}^{9803}\]

\[2S_n > 2(\sqrt{9803}-1)\]

\[S_n > \sqrt{9803}-1\]

Then applicate the left Riemann sum of $y=\frac{1}{\sqrt{x}}$

\[2S_n-1 < \int_{1}^{9801} \frac{1}{\sqrt{x}} \,dx\]

\[2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}\]

\[S_n-\frac{1}{2} > \sqrt{9801}-1\]

\[S_n < \sqrt{9801}-\frac{1}{2}\]


We conclude that:

$\sqrt{9803}-1 < S_n < \sqrt{9801}-\frac{1}{2}$

$\lfloor S_n \rfloor = \boxed{98}$

~cassphe

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