Given,

$$1 + 3 + {3^2} + {3^3} + \,\,.....\,\, + \,\,{3^{2021}}$$

$$ = {3^0} + {3^1} + {3^2} + {3^3} + \,\,....\,\, + \,\,{3^{2021}}$$

This is a G.P with common ratio = 3

$$\therefore$$ Sum $$ = {{1({3^{2022}} - 1)} \over {3 - 1}}$$

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

$$ = {{{{({3^2})}^{2011}} - 1} \over 2}$$

$$ = {{{{(10 - 1)}^{1011}} - 1} \over 2}$$

$$ = {{\left[ {{}^{1011}{C_0}\,.\,{{10}^{1011}} - {}^{1011}{C_1}\,.\,{{10}^{1010}} + \,\,.....\,\, - \,\,{}^{1011}{C_{1009}}\,.\,{{(10)}^2} + {}^{1011}{C_{1010}}\,.\,10 - {}^{1011}{C_{1011}}} \right] - 1} \over 2}$$

$$ = {{{{10}^2}\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1009}} - {}^{1011}{C_1}\,.\,(1008) + \,\,.....\,\,{}^{1011}{C_{1009}}} \right] + 10110 - 1 - 1} \over 2}$$

$$ = {{100k + 10110 - 2} \over 2}$$

$$ = {{100k + 10108} \over 2}$$

$$ = 50k + 5054$$

$$ = 50k + 50 \times 101 + 4$$

$$ = 50[k + 101] + 4$$

$$ = 50k' + 4$$

$$\therefore$$ By dividing 50 we get remainder as 4.