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Calculus Level 5

n = 0 L 2 n + 1 ( 2 n + 1 ) 2 ( 2 n n ) \large\sum _{ n=0 }^{ \infty }{ \dfrac { { L }_{ 2n+1 } }{ (2n+1)^{ 2 } \binom{2n}{n}} }

Let L n L_n denote the n n th Lucas number , where L 1 = 1 L_1 = 1 , L 2 = 3 L_2 = 3 and L n = L n 1 + L n 2 L_n = L_{n-1} + L_{n-2} for n 3 n \ge 3 .

If the series above can be expressed as π A ln ( B C D + E F G H ) + I J K , -\dfrac\pi A \ln\left( B - C\sqrt D+ E\sqrt{F - G\sqrt H} \right) + \dfrac IJ K ,

where A A , B B , C C , D D , E E , F F , H H , I I and J J are positive integers , with I I and J J being coprime integers; D D , E E and H H square-free; and K K denotes the Catalan's constant , find A + B + C + D + E + F + G + H + I + J A+B+C+D+E+F+G+H+I+J .


The answer is 11244.

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