JEE Main 20203 Sep, Morning shiftMathematicsMediumSequence and Series179× since 2002The value of (0.16)log2.5(13+132+....to ∞){\left( {0.16} \right)^{{{\log }_{2.5}}\left( {{1 \over 3} + {1 \over {{3^2}}} + ....to\,\infty } \right)}}(0.16)log2.5(31+321+....to∞) is equal to ______.Your answerType your answer. Integers and decimals are accepted.CheckSkip · show solution