Let where and are relatively prime positive integers. Find

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Let where and are relatively prime positive integers. Find

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Let $\sum_{n = 0}^{123456789} \frac{3n^2 + 9n + 7}{(n^2 + 3n + 2)^3} = \frac{a}{b},$ where $a$ and $b$ are relatively prime positive integers. Find $b-a.$

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geno3141 · Answer 1 · 2020-03-22T04:20:33

The sum is close to 1/8 ... so the value of b -a is close to 7.

Guest · Answer 2 · 2020-03-22T06:06:51

This is what Wolfram/Alpha gives as the sum which appears to be an irrational number !!!

https://www.wolframalpha.com/input/?i=%E2%88%91%5B+%283n%5E2+%2B+9n+%2B+7%29+%2F+%28n%5E2+%2B+3n+%2B2%29%5E2%2C+n%2C+0%2C+123456789+%5D

∑[ (3n^2 + 9n + 7) / (n^2 + 3n +2)^2, n, 0, 123456789 ] =3.289868109.............etc.

Alan · Answer 3 · 2020-03-23T10:46:05

As follows:

Let where and are relatively prime positive integers. Find

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Let where and are relatively prime positive integers. Find

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