MathematicsMedium64×since 2002Q2894If 12 . 310+122 . 39+ ..... + 1210 . 3=K210 . 310{1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}2.3101+22.391+.....+210.31=210.310K, then the remainder when K is divided by 6 is :A1B2C3D5Check answerSkip