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∑n=1∞1n2(2nn)=ABζ(C)\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 }\left( \begin{matrix} 2n \\ n \end{matrix} \right) } } =\frac { A }{ B } \zeta \left( C \right) n=1∑∞n2(2nn)1=BAζ(C)
In the equation above, A,B,CA,B,CA,B,C are positive integers such that AAA and BBB are coprime.
Find A+B+C.A+B+C.A+B+C.
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