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Question Number 141846 by ArielVyny last updated on 24/May/21
∑k⩾0(Cnk)2
Answered by mindispower last updated on 24/May/21
∑k⩾0Cnk.Cnk(1+x)n(1+x)n=∑k⩾0∑i⩾0CnkCnixk+i=∑2nj=0C2njxjifwecomparcoefficientof/XnC2nn=∑k⩾0∑i⩾0CnkCnj,i+j=n=∑k⩾0CnkCnn−k=C2nn⇔∑k⩾0(Cnk)2=C2nn
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