Question

1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+1/(9√7+7√9)+......+1/(49√47+47√49)

1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+1/(9√7+7√9)+......+1/(49√47+47√49)=？化简

Answer 1

1/[(2n+1)√(2n-1)+(2n-1)√(2n+1)]
=1/{[√(2n+1)(2n-1)][√(2n+1)+√(2n-1)]}
=(1/2)[√(2n+1)-√(2n-1)]/[√(2n+1)(2n-1)]
=(1/2)[1/√(2n-1)-1/√(2n+1)]

1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+1/(9√7+7√9)+......+1/(49√47+47√49)
=(1/2)[1/1-1/√3+1/√3-1/√5+...+1/√47-1/√49)
=(1/2)(1-1/√49)
=(49-√49)/98

Answer 2

每个因式上下乘以共轭，例如1/(3+√3) = (3-√3)/(9-3)=(3-√3)/6 = 1/2 -√3/6 ;又例如1/(5√3+3√5) = (5√3+3√5) / (75-45) = √3/6 - √5/10
正好可以前后相消，如此类推即可化简。你可以自己试试算算：）

Answer 3

1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+1/(9√7+7√9)+......+1/(49√47+47√49)=1/√3(√3+1)+1/√15(√5+√3)+1/√35(√7+√5)+......+1/√49*47*(√49+√47)
=(√3-1)/2√3+(√5-√3)/2√15+(√7-√5)/2√35+.....+(√49-√47)/2√49*47
=1/2*(1-√3/3+√3/3-√5/5+√5/5-√7/7+.....+√47/47-√49/49)
=1/2*(1-1/7)
=3/7