🅦🅗🅨 the resistivity of the top metal layers will be low compared to the lower metal layers eve though you use the same metal for all the layers?

Why the resistivity of the top metal layers will be low compared to the lower metal layers eve though you use the same metal for all the layers?

Theoretically, one would want to have all metal layers with low resistance. Practically, there are limitations - cost and technology - that lead to finite metal resistance .All metal layers can be made of copper, and copper has much lower resistance than aluminum (~1.7e-6 Ohm*cm vs ~2.7e-6 Ohm*cm). However copper technology is more expensive than aluminum technology, so there is a cost-performance trade-off. 

From technology viewpoint, you can make a metal layer very thick (to make sheet rho value lower), but then you can not make metal line very narrow (lateral coupling capacitance increases and results the cross talk) .So if you want to achieve very fine metal pitch (to provide high integration density - i.e. number of devices per unit area), the metal thickness can not be made very large. The solution is to use thinner (and thus more resistive) metal layers for low layers (i.e. M1, M2 ...) and for local routing, and thicker (less resistive) layers with larger width and spacing for long-range routing on the higher levels.