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∑n=01234567893n2+9n+7(n2+3n+2)3\large \sum_{n=0}^{123456789} \dfrac{3n^2+9n+7}{(n^2+3n+2)^3} n=0∑123456789(n2+3n+2)33n2+9n+7
Given that the summation above is equal to AB\dfrac ABBA, where AAA and BBB are coprime positive integers, find B−AB-AB−A.
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