Counting & ProbabilityHardBinomial Theorem, Pascal & Combinatorial Identities ShareBinomial Theorem, Pascal & Combinatorial Identities · Hard problemEvaluate ∑k=06(6k)2k\displaystyle\sum_{k=0}^{6}\binom{6}{k}2^{k}k=0∑6(k6)2k.Check Show answer Walk through it step by step Practice more like this Review this concept