If $${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :

<p>$${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$</p> <p>$$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$$</p> <p>$$ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$$</p> <p>Now $K = {(1 + 2)^{10}} - {2^{10}}$</p> <p>$$ = {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$$</p> <p>$= {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$</p> <p>$= 1 + 20 + 5120 + 6\lambda$</p> <p>$= 5136 + 6\lambda + 5$</p> <p>$= 6\mu + 5$</p> <p>$\lambda ,\,\mu \in N$</p> <p>$\therefore$ remainder = 5</p>