It is simpler than it looks!

Calculus Level 5

n=11n2(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)

In the equation above, A,B,CA,B,C are positive integers such that AA and BB are coprime.

Find A+B+C.A+B+C.

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