Compute 1≤a<b<c∑2a3b5c1.(The sum is taken over all triples (a,b,c) of positive integers such that 1≤a<b<c.)
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Let x=a,y=b−a, and z=c−b, so x≥1,y≥1, and z≥1. Also, b=a+y=x+y and c=b+z=x+y+z, so 1≤a<b<c∑2a3b5c1=x=1∑∞y=1∑∞z=1∑∞2x3x+y5x+y+z1=x=1∑∞y=1∑∞z=1∑∞30x15y5z1=x=1∑∞30x1y=1∑∞15y1z=1∑∞5z1=291⋅141⋅41=16241.