Question 329 of 347 Medium MCQ
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 :
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\[{1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}\]
\[ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}\]
\[ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}\]
Now K = ( 1 + 2 ) 10 − 2 10 K = {(1 + 2)^{10}} - {2^{10}} 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 Answer the question to unlock