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(2+4+...+2012)-(1+3+...+2011)

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Tudorpingu
Poti sa folosesti suma lui Gaus: S1=N×(N+1):2
Icecon2005
(2+4+...+2012)-(1+3+...+2011)


(2+4+...+2012)=2(1+2+3+.....+1006)=2×[(1006×1007):2]=1013042

(1+3+...+2011) =suma de numere consecutive impare=1+3+5+7+...+ (2×1005)+1 =1005²=1012036


(2+4+...+2012)-(1+3+...+2011) =1013042-1012036 =1006







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