The remainder on dividing  $1+3+3^2+3^3+\dots .+3^{2021}$ by 50 is  

$1+3+3^2+\dots \dots 3^{2021}=\frac{3^{2022}-1}{2}=\frac{9^{1011}-1}{2}$

$\Rightarrow \frac{(10-1{)}^{1011}-1}{2}=\frac{100\lambda +{}^{1011}C_{1010}\times 10-1}{2}=50\lambda +5054$
when divided by 50 gives remainder = 4

The remainder on dividing  $1+3+3^2+3^3+\dots .+3^{2021}$ by 50 is  4.