QJEEM22Mab9ec: JEE Main 2022 Mathematics - Class 11 JEEM22Mab9ec Solution: The remainder on dividing 1 + 3 + 3 2 + 3 3 + ..... + 3 2021 by …

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.