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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 \(b-a.\)

 Mar 22, 2020
 #1
avatar+21017 
0

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

 Mar 22, 2020
 #2
avatar+485 
+1

Geno how did you get the sum as close to 1/8? just asking

jfan17  Mar 22, 2020
 #3
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0

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.

 Mar 22, 2020
 #5
avatar+30088 
+2

The denominator is raised to the power 3, not 2.

Alan  Mar 23, 2020
 #4
avatar+30088 
+2

As follows:

 

 Mar 23, 2020

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