Your concept is flawed.
Because the math is so small, let's make it 1000m (cause I have the Tables for copper Ω/100m). So 1000m of #2 vs 100m of #11AWG, 100m of #10, .... to 100m of #2.
1000m of #2 has a resistance of 0.512Ω. 100m of #11 has a resistance of 0.413Ω, 100m of #10 0.327Ω. This is already greater than 1000m of #2. So your voltage drop would be greater than 3% (at 10m).
Actual math:
\begin{array} {|r|r|}\hline Length & \#2 & \#11 to \#2 & AWG \\ \hline 100 & 0.0512Ω & 0.0512Ω & \#2 \\ \hline 100 & 0.0512Ω & 0.0646Ω & \#3 \\ \hline 100 & 0.0512Ω & 0.0815Ω & \#4 \\ \hline 100 & 0.0512Ω & 0.1027Ω & \#5 \\ \hline 100 & 0.0512Ω & 0.129Ω & \#6 \\ \hline 100 & 0.0512Ω & 0.163Ω & \#7 \\ \hline 100 & 0.0512Ω & 0.206Ω & \#8 \\ \hline 100 & 0.0512Ω & 0.26Ω & \#9 \\ \hline 100 & 0.0512Ω & 327Ω & \#10 \\ \hline 100 & 0.0512Ω & 0.413Ω & \#11 \\ \hline \hline 1000m & 0.512Ω & 1.798Ω & Total \\ \hline \end{array}
So your #11 to #2 cable would have 351% more resistance.
To achive #2 equivalency, you can't start (end) with #2, but #2 must be somewhere in the middle.
Now same approach but going from 4/0 to #6 will get the resistance down to 0.5636Ω. This is bigger than 0.512Ω, but it illustrates the problem.
\begin{array} {|r|r|}\hline AWG & Ω/100m & Area (CM) & Area (CM) \\ \hline 4/0 & 0.016Ω & 211,600 & 66,373 \\ \hline 3/0 & 0.02Ω & 167,810 & 66,373 \\ \hline 2/0 & 0.026Ω & 133,080 & 66,373 \\ \hline 1/0 & 0.032Ω & 105,530 & 66,373 \\ \hline \#1 & 0.0406Ω & 83,694 & 66,373 \\ \hline \#2 & 0.0512Ω & 66,373 & 66,373 \\ \hline \#3 & 0.0646Ω & 52,634 & 66,373 \\ \hline \#4 & 0.0815Ω & 41,742 & 66,373 \\ \hline \#5 & 0.1027Ω & 33,102 & 66,373 \\ \hline \#6 & 0.129Ω & 26,250 & 66,373 \\ \hline Total & 0.5636Ω & 921,815CM & 663,730CM \\ \hline \end{array}
So this approach uses 38.8% more copper. Aside from cost of manufacturing, more copper means more cost.
So more resistance or greater cost. Either way, your concept is flawed.