If the value of
$${\left( {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....upto\,\infty } \right)^{{{\log }_{(0.25)}}\left( {{1 \over 3} + {1 \over {{3^2}}} + {1 \over {{3^3}}} + ....upto\,\infty } \right)}}$$
is $l$, then $l$2 is equal to _______________.
Answer (integer)
3
Solution
$$l = {\left( {\underbrace {1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}}}_S + ....} \right)^{{{\log }_{0.25}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ...} \right)}}$$<br><br>$S = 1 + {2 \over 3} + {6 \over {{3^2}}} + {{10} \over {{3^3}}} + ....$<br><br>${S \over 3} = {1 \over 3} + {2 \over {{3^2}}} + {6 \over {{3^3}}} + .....$${{2x} \over 3} = 1 + {1 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + ....$<br><br>${{2S} \over 3} = {4 \over 3} + {4 \over {{3^2}}} + {4 \over {{3^3}}} + .....$<br><br>$S = {3 \over 2}\left( {{{4/3} \over {1 - 1/3}}} \right) = 3$<br><br>Now, $$l = {\left( 3 \right)^{{{\log }_{0.25}}\left( {{{1/3} \over {1 - 1/3}}} \right)}}$$<br><br>$$l = {3^{{{\log }_{\left( {(1/4)} \right)}}\left( {{1 \over 2}} \right)}} = {3^{1/2}} = \sqrt 3 $$<br><br>$\Rightarrow$ l<sup>2</sup> = 3
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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