4, 3, 2, 1: Part 2

Calculus Level 4

n=1n243nn!=eab\large \sum _{ n=1 }^{ \infty }{ \dfrac { { n }^{ 2 } }{ 4\cdot { 3 }^{ n }\cdot n! } } =\dfrac { \sqrt [ a ]{ e } }{ b }

If the equation above holds true for positive integers, find a+ba+b.

Clarification: e2.71828e \approx 2.71828 denotes the Euler's number.

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