MathematicsMedium178×since 2002Q5305The sum ∑k=120k12k\sum\limits_{k = 1}^{20} {k{1 \over {{2^k}}}}k=1∑20k2k1 is equal toA2−112192 - {11 \over {{2^{19}}}}2−21911B2−32172 - {3 \over {{2^{17}}}}2−2173C1−112201 - {11 \over {{2^{20}}}}1−22011D2−212202 - {21 \over {{2^{20}}}}2−22021Check answerSkip