1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+…+1/(49√47+47√49)=

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/11 16:31:50
1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+…+1/(49√47+47√49)=

1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+…+1/(49√47+47√49)=
1/(3+√3)+1/(5√3+3√5)+1/(7√5+5√7)+…+1/(49√47+47√49)=

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

1/(3+√3)
=(3-√3)/(3+√3)(3-√3)
=(3-√3)/6
=1/2(1-1/√3)
其它同理
原式
=(1/2)[1/1-1/√3+1/√3-1/√5+...+1/√47-1/√49)
=(1/2)(1-1/√49)
=(49-√49)/98