Inspired by Mrigank

Calculus Level 3

⌊ 100 99 + 400 399 + 900 899 + ⋯ + 407232400 407232399 ⌋ = a b \left\lfloor \dfrac{100}{99} + \dfrac{400}{399} + \dfrac{900}{899} + \cdots +\dfrac{407232400}{407232399}\right \rfloor = ab

The equation above holds true for positive coprime integers a a and b b . Find the value of a + b a+b .

Note : ⌊ ⋅ ⌋ \left\lfloor \cdot \right\rfloor denotes the floor function .

Inspiration: Find the product: the answer is product


The answer is 1011.

Naren Bhandari by Naren Bhandari , 21, India

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2 solutions

Chew-Seong Cheong
Apr 15, 2018

S = 100 99 + 400 399 + 900 899 + ⋯ + 407232400 407232399 = ∑ k = 1 2018 100 k 2 100 k 2 − 1 = ∑ k = 1 2018 ( 1 + 1 100 k 2 − 1 ) = 2018 + ∑ k = 1 2018 1 100 k 2 − 1 \begin{aligned} S & = \frac {100}{99} + \frac {400}{399} + \frac {900}{899} + \cdots + \frac {407232400}{407232399} \\ & = \sum_{k=1}^{2018} \frac {100k^2}{100k^2-1} \\ & = \sum_{k=1}^{2018} \left(1+\frac 1{100k^2-1}\right) \\ & = 2018 + \sum_{k=1}^{2018} \frac 1{100k^2-1} \end{aligned}

We note that ∑ k = 1 2018 1 100 k 2 − 1 \displaystyle \sum_{k=1}^{2018} \frac 1{100k^2-1} is convex and that

∑ k = 1 2018 1 100 k 2 − 1 < ∫ 0.5 2018.5 1 100 x 2 − 1 d x = 1 2 ∫ 0.5 2018.5 ( 1 10 x − 1 − 1 10 x + 1 ) d x = 1 20 ∣ ln ⁡ ( 10 x − 1 10 x + 1 ) ∣ 0.5 2018.5 ≈ 0.0203 \begin{aligned} \sum_{k=1}^{2018}\frac 1{100k^2-1} & < \int_{0.5}^{2018.5} \frac 1{100x^2-1} \ dx \\ & = \frac 12 \int_{0.5}^{2018.5} \left(\frac 1{10x-1} - \frac 1{10x+1}\right) \ dx \\ & = \frac 1{20} \left|\ln \left(\frac {10x-1}{10x+1} \right)\right|_{0.5}^{2018.5} \\ & \approx 0.0203 \end{aligned}

Therefore, ⌊ S ⌋ = 2018 \lfloor S \rfloor = \boxed{2018} .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

UPVOTED SIR , It's really awesome . Thank your for the solution. I hope you enjoyed the problem. :)

n =2018 in the fourth line. Is that you mean ?

Naren Bhandari - 3 years, 1 month ago
Naren Bhandari
Apr 10, 2018

Since 100 99 + 400 399 + ⋯ + 4072232400 4072232399 = ∑ k = 1 2018 100 k 2 100 k 2 − 1 = ∑ k = 1 2018 ( 1 + 1 2 ( 1 10 k − 1 − 1 10 k + 1 ) ) \dfrac{100}{99}+\dfrac{400}{399} + \cdots + \dfrac{4072232400}{4072232399} = \sum_{k=1}^{2018} \dfrac{100k^2}{100k^2-1} = \sum_{k=1}^{2018} \left(1+\dfrac{1}{2}\left(\dfrac{1}{10k-1} -\dfrac{1}{10k+1}\right)\right) ∴ ⌊ ∑ k = 1 2018 ( 1 + 1 2 ( 1 10 k + 1 − 1 10 k − 1 ) ) ⌋ = 2018 = 2 ⋅ 1009 \therefore \left\lfloor \sum_{k=1}^{2018} \left(1+\dfrac{1}{2}\left(\dfrac{1}{10k+1} -\dfrac{1}{10k-1}\right)\right)\right\rfloor = 2018 = 2\cdot 1009 Therefore, a + b = 2 + 1009 = 1011 a+b =2+1009 = 1011 .

1 Helpful 0 Interesting 0 Brilliant 0 Confused

thank you for this good question; can you please change the title "Inspired by Mringank" to Inspired by Mrigank"

Mrigank Shekhar Pathak - 3 years, 2 months ago

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Your welcome. :) Sorry for misspelled. Now, I have fixed it.

I wrote solution for the inspired problem of yours with the same approach and I could concluded it converges between 1 to 2 but couldn't complete it so I deleted it.

Naren Bhandari - 3 years, 2 months ago

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Please look at the solution of "Find the Product: the answer is a product" I have posted.

Mrigank Shekhar Pathak - 3 years, 2 months ago

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