Calculati :
a)1+2+3+...+100
b)3+6+9+12+...+57
c)1999+1999*2000-2001*1997
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[tex]a)~1+2+...+100= \frac{100*(100+1)}{2} =50*101=\boxed{5050}.\\\\
b)3+6+9+...+57=3*1+3*2+3*3+...+3*19\\3*(1+2+...+19)=3* \frac{19*(19+1)}{2} =3*19*10=3*190=\boxed{570}.\\\\
c) 1999+1999*2000-2001*1997=1999*(1+2000)-2001*1997\\=1999*2001-2001*1997=2001*(1999-1997)=2001*2=\boxed{4002}[/tex]
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