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vartosusimonamihaela
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1+2+3+.....+ 1199+1200 rezolvare completa va rog

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iondaniela

1+2+3+.....+1199+1200= =1200*(1200+1):2= =1200*1201:2= =600*1201=720600

margacatalin

[latex]\displaystyle \boxed{ \frac{n(n+1)}{2} } \\\\ \text{n=ultimul numar din sir } \\\\ 1+2+3+.....+ 1199+1200= \frac{1200(1200+1)}{2}= \\ \\ \\ \frac{\not1200\cdot1201}{\not2}= 600 \cdot 1201 \\ \\ \\ 600\cdot1201=720600[/latex]

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