Difference between revisions of "2018 MPFG Problem 19"

Revision as of 09:01, 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 $y=\frac{1}{\sqrt{x}}$

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

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

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

$S_n > \sqrt{9803}-1$

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

@@ Line 15: / Line 15: @@
 [[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>
@@ Line 27: / Line 27: @@
 [[File:Left_rie.jpg|750px|center]]
-<cmath>2S_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>2S_n-1 < \left. (2x^{\frac{1}{2}})\right|_{1}^{9801}</cmath>

Page

Toolbox

Search

Difference between revisions of "2018 MPFG Problem 19"

Revision as of 09:01, 24 August 2025

Problem 19

Solution 1