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求 [cot(π/12)]^2003+[tan(π/12)]^2003 除以 9 的余数
(2+√3)^2003+(2-√3)^2003
=∑(2003Cr)[2^(2003-r)](√3^r)[1+(-1)^r],r=0,1,...,2003
=∑(2003C2r)[2^(2004-2r)](3^r),r=0,1,...,1001
{∑(2003C2r)[2^(2004-2r)](3^r),r=0,1,...,1001}rem9
={∑(2003C2r)[2^(2004-2r)](3^r),r=0,1}rem9
=[2^2004+(2003C2)(2^2002)(3)]rem9
=[8^668+(2003)(1001)(2)(8^667)(3)]rem9
=[1-(2003)(6006)]rem9
=[1-(5)(3)]rem9
=4 |
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