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Question Number 136098 by SOMEDAVONG last updated on 18/Mar/21
I).compute∑1000k=0C20002k.
Answered by Olaf last updated on 18/Mar/21
(1−1)2000=0=∑2000p=0(−1)pCp20000=∑1000k=0(−1)2kC2k2000+∑999k=0(−1)2k+1C2k+120000=∑1000k=0C2k2000−∑999k=0C2k+12000∑1000k=0C2k2000=∑999k=0C2k+12000(1+1)2000=22000=∑2000p=0Cp200022000=∑1000k=0C2k2000+∑999k=0C2k+1200022000=∑1000k=0C2k2000+∑1000k=0C2k200022000=2∑1000k=0C2k2000∑1000k=0C2k2000=12(22000)=21999
Commented by SOMEDAVONG last updated on 18/Mar/21
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