n=1∑∞(2n4n)−1=X+Yπ−ZZ2ln(21+Z)
The equation above holds true for rational numbers X, Y, and Z. Find XYZ.
Note: (⋅⋅) is the binomial coefficient. The first few terms of the series are as follows:
n=1∑∞(2n4n)−1=(24)1+(48)1+(612)1+⋯=61+701+9241+⋯.