数列求和问题
数列相消法的常见的拆项公式: 1/n(n+1)=( ) 1/(2n-1)(2n+1)= ( ) 1/[根号下n+根号下(n+1)]= ( )
1/n(n+1)=(1/n)-[1/(n+1)] 1/(2n-1)(2n+1)= (1/2)[1/(2n-1)]- [1/((2n+1)] 1/[根号下n+根号下(n+1)]=根号下(n+1)]-根号下n
答:设An=2cos(an),0≤an≤π/2,a1=π/4, An+1=√(2+ An)=(√2)√(1+ cos(an))= 2cos(an/2)=2cos(...详情>>
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