Here's my solution (click to reveal):-
Because x!(x+1)!=(x!)2(x+1)
It follows that x=1∏100x! =y=1∏50((2y−1)!(2y)!) =(y=1∏50(2y−1)!)2×y=1∏50(2y) =(y=1∏50(2y−1)!)2×250×50! =(225×y=1∏50(2y−1)!)2×50!
The rest is child's play
Here's my solution (click to reveal):-
Because x!(x+1)!=(x!)2(x+1)
It follows that x=1∏100x! =y=1∏50((2y−1)!(2y)!) =(y=1∏50(2y−1)!)2×y=1∏50(2y) =(y=1∏50(2y−1)!)2×250×50! =(225×y=1∏50(2y−1)!)2×50!
The rest is child's play
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