2018 MPFG Problem 19: Difference between revisions

Revision as of 07:55, 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 $y=\frac{1}{\sqrt{x}}$

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

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

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

$S_n > \sqrt{9803}-1$

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

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

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

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

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

We conclude that:

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

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

~cassphe

Revision as of 05:15, 24 August 2025 view source Cassphe (talk \| contribs) editor 207 edits →Solution 1 ← Older edit		Revision as of 07:55, 24 August 2025 view source Cassphe (talk \| contribs) editor 207 edits →Solution 1 Newer edit →
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	<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}-1/2</cmath>		<cmath>S_n < \sqrt{9801}-\frac{1}{2}</cmath>

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2018 MPFG Problem 19: Difference between revisions

Revision as of 07:55, 24 August 2025

Problem 19

Solution 1