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va rog ajutor 2003+2.(1+2+...+2002)

Răspuns :

2003+2x(2003x2002/2)= 2003+2x2005003= 2003+401006=403009
Stefiandrea3
2003+2*(1+2+...+2002)=
2003+2* 2005003=
2003+4010006=
4012009

1+2+...+2002=[tex] \frac{n*(n+1)}{2} [/tex]
1+2+...+2002=[tex] \frac{2002*(2002+1)}{2} [/tex] ⇒ [tex] \frac{4010006}{2} = 2005003[/tex]


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