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Question Number 119750 by 675480065 last updated on 26/Oct/20
∏1019k=1[2k2k−1]=?
Answered by Bird last updated on 26/Oct/20
∏k=11019[2k2k−1]=∏k=11019[2k−1+12k−1]=∏k=11019[1+12k−1]=∏k=11019(1+[12k−1])=2∏k=21019(1+[12k−1])but0<12k−1<1fork∈[[2,1019]]⇒[12k−1]=0⇒∏k=11019[2k2k−1]=2
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