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Va rog ajutati-ma si pe mine urgent1+2+3+...1000= ;1+2+3+...2001=;1+2+3+...50;2+4+6+...2004 va rog mult

Răspuns :

[tex]\displaystyle a).1+2+3+...+1000= \frac{1000(1000+1)}{2} = \frac{1000 \cdot 1001}{2} = \\ \\ = \frac{1001000}{2} =500500 \\ \\ b).1+2+3+...+2001= \frac{2001(2001+1)}{2} = \frac{2001 \cdot 2002}{2} = \\ \\ = \frac{4006002}{2} =2003001 \\ \\ c).1+2+3+...+50= \frac{50(50+1)}{2} = \frac{50 \cdot 51}{2} = \frac{2550}{2} =1275 \\ \\ d).\displaystyle 2+4+6+...+2004=2(1+2+3+...+1002)= \\ \\ =2 \cdot \frac{1002(1002+1)}{2} =2 \cdot \frac{1002 \cdot 1003}{2} =\not 2 \cdot \frac{1005006}{\not2} =1005006[/tex]

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