[tex]\it f(1)+f(2)+f(3)+\ ...\ +f(50)=\underbrace{\it(2+1)+(4+1)+(6+1)+\ ...\ +(100+1)}_{50\ paranteze}=\\ \\ \\ =(2+4+6+\ ...\ +100)+\underbrace{\it1+1+1+\ ...\ +1}_{50\ termeni}=2\cdot(1+2+3+\ ...\ +50)+\\ \\ \\ +50\cdot1=\not2\cdot\dfrac{50\cdot51}{\not2}+50=2550+50=2600[/tex]
