/sci/ - Science & Math » Thread #10875321

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Downtime Math #4

Anonymous Fri 09 Aug 06:27:43 2019 No.10875321 View Reply Original Report

Quoted By: >>10875328 >>10875574

Hi /sci/
It seems like the last problem was way too easy.
Hopefully this one's a bit more challenging.
Last post: >>10873690

Anonymous

Anonymous Fri 09 Aug 2019 06:35:36 No.10875328 Report

Quoted By: >>10875382

>>10875321
This is a simple telescoping sum
$\sum_{n=1}^{2019} \frac{n^2 + n + 1}{(n+1)^5 - n^5 - 1} \\<br />= \sum_{n=1}^{2019} \left(\frac{1}{5n} - \frac{1}{5(n+1)} \right) \\<br />= \frac{1}{5} - \frac{1}{5(2020)} \\<br />= \frac{2019}{10100}<br />$

Anonymous

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Anonymous Fri 09 Aug 2019 07:14:51 No.10875382 Report

Quoted By: >>10875396 >>10875622

>>10875328
Damn that was quick.
Here's the solution.

Anonymous

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Anonymous Fri 09 Aug 2019 07:25:28 No.10875396 Report

Quoted By: >>10875411

>>10875382
I am a passer by, can you explain how to get from the first line to the second?
I feel like I am missing something fundamental and crucial.

Anonymous

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Anonymous Fri 09 Aug 2019 07:40:35 No.10875411 Report

Quoted By:

>>10875396
It's just expanding and simplifying.
Nothing to worry about.

Anonymous

Anonymous Fri 09 Aug 2019 11:08:16 No.10875574 Report

Quoted By: >>10875622

>>10875321
The denominator factorizes into 5n*(n+1)*(n^2 + n + 1), and the numerator is n^2 + n + 1. So, the 1/n partial sum expansion works here, so its 1/5 - 1/10100

Anonymous

Anonymous Fri 09 Aug 2019 11:46:37 No.10875622 Report

Quoted By:

>>10875574
That's correct.
Refer to >>10875382 for full solution.

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