The remainder on dividing 1 + 3 + 3² + 3³ + ..... + 3²⁰²¹ by 50 is _________.

<p>Given,</p> <p>$1 + 3 + {3^2} + {3^3} + \,\,.....\,\, + \,\,{3^{2021}}$</p> <p>$= {3^0} + {3^1} + {3^2} + {3^3} + \,\,....\,\, + \,\,{3^{2021}}$</p> <p>This is a G.P with common ratio = 3</p> <p>$\therefore$ Sum $= {{1({3^{2022}} - 1)} \over {3 - 1}}$</p> <p>$= {{{3^{2022}} - 1} \over 2}$</p> <p>$= {{{{({3^2})}^{2011}} - 1} \over 2}$</p> <p>$= {{{{(10 - 1)}^{1011}} - 1} \over 2}$</p> <p>$$ = {{\left[ {{}^{1011}{C_0}\,.\,{{10}^{1011}} - {}^{1011}{C_1}\,.\,{{10}^{1010}} + \,\,.....\,\, - \,\,{}^{1011}{C_{1009}}\,.\,{{(10)}^2} + {}^{1011}{C_{1010}}\,.\,10 - {}^{1011}{C_{1011}}} \right] - 1} \over 2}$$</p> <p>$$ = {{{{10}^2}\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1009}} - {}^{1011}{C_1}\,.\,(1008) + \,\,.....\,\,{}^{1011}{C_{1009}}} \right] + 10110 - 1 - 1} \over 2}$$</p> <p>$= {{100k + 10110 - 2} \over 2}$</p> <p>$= {{100k + 10108} \over 2}$</p> <p>$= 50k + 5054$</p> <p>$= 50k + 50 \times 101 + 4$</p> <p>$= 50[k + 101] + 4$</p> <p>$= 50k' + 4$</p> <p>$\therefore$ By dividing 50 we get remainder as 4.</p>