Difference between revisions of "2018 MPFG Problem 19"

Revision as of 06:03, 24 August 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.

$_n = 1/\sqrt{1}+1/\sqrt{3}+ ... + 1/\sqrt{9801}$

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

[insert pic]

$2S_n > \int{f_{1}^{9803}(\frac{1}{\sqrt{x}})}dx$

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

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

$S_n > \sqrt{9803}-1$

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

[insert pic2]

$2S_n-1 < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx$

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

$S_n-\frac{1}{2} > \sqrt{9801}-1$

$S_n < \sqrt{9801}-1/2$

We conclude that:

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

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

~cassphe

@@ Line 27: / Line 27: @@
 [insert pic2]
-<cmath>2(S_n-1) < \int{f_{1}^{9801}(\frac{1}{\sqrt{x}})}dx</cmath>
+<cmath>2S_n-1 < \int{f_{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}-1/2</cmath>
 We conclude that:
-<math>\sqrt{9803}-1 < S_n < \sqrt{9801}</math>
+<math>\sqrt{9803}-1 < S_n < \sqrt{9801}-\frac{1}{2}</math>
 <math>\lfloor S_n \rfloor = \boxed{98}</math>
 ~cassphe

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Difference between revisions of "2018 MPFG Problem 19"

Revision as of 06:03, 24 August 2025

Problem 19

Solution 1