因为nπ/(2k+1)+(2k+1-n)π/(2k+1)=π
所以cos(nπ/(2k+1))+cos((2k+1-n)π/(2k+1))=0
对于n=1到k都成立,相加之后仍然成立
所以上式成立
求证:cos(π/2k+1)+cos(2π/2k+1)+…+cos(2k-1)π/2k+1+cos2kπ/2k+1=0
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