QJEEM22Ma13ca: JEE Main 2022 Mathematics - Class 11 JEEM22Ma13ca Solution: If , then the remainder when K is divided by 6 is :

${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$\Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$

$= {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$

Now $K = {(1 + 2)^{10}} - {2^{10}}$

$= {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$

$= {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$

$= 1 + 20 + 5120 + 6\lambda$

$= 5136 + 6\lambda + 5$

$= 6\mu + 5$

$\lambda ,\,\mu \in N$

$\therefore$ remainder = 5