分母有理化
原式=(√3-1)/2+(√5-√3)/2+(√7-√5)/2+……+[√(2n+1)-√(2n-1)]/2
=[√3-1+√5-√3+√7-√5+……+√(2n+1)-√(2n-1)]/2
=[√(2n+1)-1]/2
1/√3+1+1/√5+√3+1/√7+√5+至=1/√2n+1+√2n-1.化简
1个回答
相关问题
-
化简1/(2*5)+1/(5*8)+1/(8*11)+…+1/[（3n-1）*（3n+2）]=
-
1/1*3+1/3*5+1/5*7+.+1/（2n-1）(2n+1)=n/(2n+1)
-
化简：1/根号3+1+1/根号5+根号3+1/根号7+根号5+..+1/根号2n+1+根号2n-1大神们帮帮忙
-
1/(√3 +1) +( 1/√5+√3 )+ (1/√7+√5)+.+{1/√(2n+1)+√(2n-1)}
-
化简1/1*2+1/2*3+1/3*4+.1/n(n+1)
-
1.1/1*2+1/2*3+1/3*4+……+1/n(n+1) 2.1/1*3+1/3*5+1/5*7+……+1/(2n
-
化简：1*1!+2*2!+3*3!+.+n*n!
-
1/1*5+1/3*7+1/5*9+.+1/(2n-1)(2n+3)
-
求和1/1·3+1/3·5+1/5·7+...+1/(2n+1)(2n-1)
-
求和:1/(1*3)+1/(3*5)+1/(5*7)+...+1/(2n-1)(2n+1)=?