The remainder on dividing 1 + 3 + 3 2 + 3 3 + … + 3 2021 by 50 is _____.

Let $S=1+3+3^2+\cdots ·3^{2021}$

$=\frac{3^{2022}-1}{3-1}=\frac{3^{2022}-1}{2}$

Simplifying by using binomial expansion we get$\frac{{\left(3^2\right)}^{1011}-1}{2}=\frac{(10-1{)}^{1011}-1}{2}$

$\Rightarrow \frac{{}^{1011}{C}_0{10}^{1011}{-}^{1011}{C}_1{10}^{1010}+\dots +{}^{1011}C_{1010}{10}^1-{}^{1011}C_{1011}\times 1-1}{2}$

$\frac{100q+1011\times 10-2}{2}$

$\Rightarrow \frac{100q+10110-2}{2}=\frac{100q+10108}{2}=50q+5054$

Now dividing $50q+5054$ by $50$

We get $4$ as remainder.

Hence option B is correct.