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(1+3+5+7+...+2001+2003)-(2+4+6+8+...+2000+2002)

Răspuns :

Alexandruionuttone

(1+3+5+7+…+2001+2003)-(2+4+6+8+…2000+2002)

[(1+2+3+…+2003)-2+4+6+…+2002]-2(1+2+3+…+1001)

(2003*2004:2-2*1001*1002:2)-2*1001*1002:2

(2003*1002-1001*1002)-1001*1002

2003*1002-2*1001*1002

1002(2003+2002)

1002*4005

4013010

Răspuns:

1+3+5+7+...+2001+2003=2(0+1+2+3+...+1001)+(1+1+1+1+...+1)=(2×1001×1002) /2+1002=1003002+1002

2+4+6+8+...+2000+2002=

2(1+2+3+4+...+1000+1001)=2×[1001× (1001+1)/2] =2×1001×1002 /2=1003002

Rezultatul

(1+3+5+7+...+2001+2003)-(2+4+6+8+...+2000+2002)=(1003002+1002)-1003002=1002

Explicație pas cu pas:

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