Consider a gross oversimplification of the case of increasing from two blades to four blades, same diameter:Why is that? I would have thought it would be the other way around.
Twice the blades, half the load carried by each. Half the load, half the downwash angle and half the size of the "resultant vector" --> 1/4 the induced drag per blade. But there are now twice as many blades so 1/4 x 2 = 1/2 the induced drag.