Let S=1+3+32+⋯⋅32021
=3−132022−1=232022−1
Simplifying by using binomial expansion we get2(32)1011−1=2(10−1)1011−1
⇒21011C0101011−1011C1101010+…+C10101011101−C10111011×1−1
2100q+1011×10−2
⇒2100q+10110−2=2100q+10108=50q+5054
Now dividing 50q+5054 by 50
We get 4 as remainder.
Hence option B is correct.