JEE Main 20198 Apr, Evening shiftMathematicsMediumSequence and Series179× since 2002The sum ∑k=120k12k\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}}k=1∑20k2k1 is equal toAOption A. 2−112192 - {11 \over {{2^{19}}}}2−21911BOption B. 2−32172 - {3 \over {{2^{17}}}}2−2173COption C. 1−112201 - {11 \over {{2^{20}}}}1−22011DOption D. 2−212202 - {21 \over {{2^{20}}}}2−22021CheckSkip · show solution