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...