Question

ajutor! ma puteti ajuta?

Answer 1

$Pentru~a~calcula~suma,~trebuie~sa~ne~folosim~de~punctul~anterior. \\ \\ a) ~\frac{1}{x(x+2)}= \frac{ \frac{(x+2)-x}{2} }{x(x+2)}= \frac{1}{2}* \frac{(x+2)-x}{x(x+2)}= \frac{1}{2}* [\frac{x+2}{x(x+2)}- \frac{x}{x(x+2)}]= \\ = \frac{1}{2}*( \frac{1}{x}- \frac{1}{x+2}). \\ \\ b)~ \frac{1}{1*3}= \frac{1}{2}(1- \frac{1}{3}) \\ ~~~~ \frac{1}{3*5}= \frac{1}{2}( \frac{1}{3}- \frac{1}{5}) \\ ~~~~ \frac{1}{5*7}= \frac{1}{2}( \frac{1}{5}- \frac{1}{7}) \\ ~~~~........................$
$\frac{1}{31*33}= \frac{1}{2}( \frac{1}{31}- \frac{1}{33}) ~~~~(adunam~relatiile). \\ \\ S= \frac{1}{2}( 1-\frac{1}{3}+ \frac{1}{3}- \frac{1}{5}+ \frac{1}{5}- \frac{1}{7}+ ...+ \frac{1}{31}- \frac{1}{33})= \\ = \frac{1}{2}(1- \frac{1}{33} )= \\ = \frac{1}{2} * \frac{32}{33}= \\ = \frac{16}{33}.$