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∑n=1∞n24⋅3n⋅n!=eab\large \sum _{ n=1 }^{ \infty }{ \dfrac { { n }^{ 2 } }{ 4\cdot { 3 }^{ n }\cdot n! } } =\dfrac { \sqrt [ a ]{ e } }{ b } n=1∑∞4⋅3n⋅n!n2=bae
If the equation above holds true for positive integers, find a+ba+ba+b.
Clarification: e≈2.71828e \approx 2.71828e≈2.71828 denotes the Euler's number.
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