[tex]\it \bigg(1+\dfrac{3}{2}+\dfrac{4}{3}+\ ...\ +\dfrac{100}{99}\bigg)-\bigg(\dfrac{1}{2}+\dfrac{1}{3}+\ ...\ +\dfrac{1}{99}\bigg)=\\ \\ \\ =1+\underbrace{\bigg(\dfrac{3}{2}-\dfrac{1}{2}\bigg)+\bigg(\dfrac{4}{3}-\dfrac{1}{3}\bigg)+\ ...\ +\bigg(\dfrac{100}{99}-\dfrac{1}{99}\bigg)}_{98\ de\ paranteze}=\underbrace{1+1+1+ ... +1}_{de\ 99\ ori}=99[/tex]
