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∑n=1252n−1n(n+1)(n+2) \large \displaystyle\sum_{n=1}^{25}\frac{2n-1}{n(n+1)(n+2)} n=1∑25n(n+1)(n+2)2n−1
If the summation above equals to ME\frac ME EM for coprime positive coprime integer, then find the value of M+EM+EM+E.
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