[tex]S=1+3+5+7+...+201= \\ =2+4+6+8+...+202 -1-1-1-...-1 = \\ 2(1+2+3+4+..+101) - 101= \\ =2 *\frac{101*102}{2} - 101 = 101*102-101 = 10302-101 = 10201[/tex]
Ca sa aflii de cate ori e -1 imparti pe 201 la 2 care e 100 rest 1, si va fi 101.
Poti aplica si formula asta direct: [tex]1+3+5+..+n =(\frac{n+1}{2})^{2}[/tex]
