IOQM 2021 P18

Art of Problem Solving

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

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IOQM 2021 P18

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562 posts

Dr_Vex

#1 h Jan 17, 2021, 6:55 AM

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If
$\sum_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right) = a + \frac {b}{c}$ where $a, b, c \in \mathbb{N}, b < c, gcd(b,c) =1 ,$ then what is the value of $a+ b ?$

This post has been edited 3 times. Last edited by Dr_Vex, Feb 15, 2021, 2:20 PM

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980 posts

#4 h Jan 17, 2021, 7:07 AM

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Dr_Vex wrote:

If
$\Sigma_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right) = a + \frac {b}{c}$ where $a, b, c \in \mathbb{N}, b < c, gcd(b,c) =1 ,$ then what is the value of $a+ b + c$

Question had $a+b$ and answer was $80$

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#5 h Jan 17, 2021, 7:09 AM

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$1 + \frac{1}{k^{2}} + \frac{1}{(k+1)^2} = \frac{k^4+2k^3+3k^2+2k+1}{k^2(k+1)^2} =\left( \frac{k^2+k+1}{k(k+1)} \right)^2$

Hence $\sum_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right) = \sum_{k=1}^{40}\frac{k^2+k+1}{k(k+1)} = \sum_{k=1}^{40}\left(1+\frac{1}{k}-\frac{1}{k+1}\right) = 40 + \frac{40}{41}$
Hence $a+b+c = \boxed{121}$

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N1RAV

#6 h Jan 17, 2021, 7:23 AM • 4 Y

Y by angle-bisector, Mango247, Mango247, Mango247

I think we had to find only $a+b$ and not $a+b+c$

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L567

#7 h Jan 17, 2021, 7:55 AM

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Yeah, so answer is just 80

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Quantum_fluctuations

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Quantum_fluctuations

#8 h Jan 17, 2021, 8:56 AM • 1 Y

Y by Mango247

Dr_Vex wrote:

If
$\Sigma_{k=1}^{40} \left( \sqrt{1 + \frac{1}{k^{2}} + \frac{1}{(k + 1)^{2}}}\right) = a + \frac {b}{c}$ where $a, b, c \in \mathbb{N}, b < c, gcd(b,c) =1 ,$ then what is the value of $a+ b ?$

This is an old problem from chinese national high school math league.

This post has been edited 1 time. Last edited by Quantum_fluctuations, Jan 17, 2021, 8:57 AM

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pog

#9 h Mar 14, 2021, 2:54 AM • 3 Y

Y by Mango247, Mango247, Mango247

It's also USAMTS 3/4/11.

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#10 h Apr 24, 2021, 6:50 PM

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It was also asked in Common Admission Test (CAT) in 2008.

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