Question

√(lung)2003+2*(1+2+...+2002)=?
P.S radicalul lung acopera tot exercitiul
multumesc!

Answer 1

√2003+2*(1+2+3+....+2002)=
=√2003+2*2002*2003:2=
=√2003+2002*2003=
=√2003*(1+2002)=
=√2003*2003=2003

Answer 2

[tex] \sqrt{2003+2(1+2+3+...+2002)}=\\ Avem~o~suma~Gauss,o~vom~calcula~folosind~formula[n(n+1)]:2\\ \sqrt{2003+2( \frac{2002*2003}{2} )} =\ \textgreater \ \sqrt{2003+2*2005003}\\ =\ \textgreater \ \sqrt{ 4010006+2003}=\ \textgreater \ \sqrt{ 4012009}=\ \textgreater \ 2003[/tex]