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∑n=146(n3−n2−3n−4)(n−1)! \large \sum_{n=1}^{46} \frac {(n^3-n^2-3n-4)}{(n-1)!} n=1∑46(n−1)!(n3−n2−3n−4)
If the summation above equals to SSS, evaluate (46!S−1)13 (46!S-1)^{\frac{1}{3}} (46!S−1)31.
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