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Let ak=(k2+1)k!a_k = (k^2 +1) k!ak=(k2+1)k! and bn=∑k=1nakb_n = \displaystyle \sum_{k=1}^{n} a_kbn=k=1∑nak.
a100b100\dfrac{a_{100}}{b_{100}}b100a100 can be represented as AB\dfrac ABBA for positive coprime integers AAA and BBB.
Find B−AB-AB−A.
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