[tex]\it E(n)=\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}\\ \\ \\ E(2)+E(3)+\ . . . \ + E(n)=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ ...\ +\dfrac{1}{n}-\dfrac{1}{n+1}=\\ \\ \\ =\dfrac{1}{2}-\dfrac{1}{n+1}=\dfrac{n-1}{2n+2}\\ \\ \\ Egalitatea\ din\ enun\c{\it t}\ devine:\\ \\ \\ \dfrac{n-1}{2n+2}=\dfrac{50}{101} \Rightarrow101n-101=100n+100 \Rightarrow 101n-100n=100+101 \Rightarrow\\ \\ \\ \Rightarrow n=201[/tex]
