$${6 \over {{3^{12}}}} + {{10} \over {{3^{11}}}} + {{20} \over {{3^{10}}}} + {{40} \over {{3^9}}} + \,\,...\,\, + \,\,{{10240} \over 3} = {2^n}\,.\,m$$, where m is odd, then m . n is equal to ____________.
Answer (integer)
12
Solution
<p>$${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$$</p>
<p>$$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$$</p>
<p>$\Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$</p>
<p>$\Rightarrow {2^n}\,.\,m = {2^{12}}$</p>
<p>$\Rightarrow m = 1$ and $n = 12$</p>
<p>$m\,.\,n = 12$</p>
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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