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∑n=3∞1(n3)=1(33)+1(43)+1(53)+1(63)+⋯= ?\displaystyle \sum_{n=3}^{\infty} \dfrac{1}{\binom{n}{3}} = \dfrac{1}{\binom{3}{3}} + \dfrac{1}{\binom{4}{3}} + \dfrac{1}{\binom{5}{3}} + \dfrac{1}{\binom{6}{3}} + \cdots = \, ? n=3∑∞(3n)1=(33)1+(34)1+(35)1+(36)1+⋯=?
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