Question

Sa se rezolve ecuatia x+3x+5x+...+13x=-490,stiind ca x apartine lui Z.
VA ROG, E URGENT!!!!!!!!!

Answer 1

$x(1+3+5+7+9+11+13)=49x=-490 =\ \textgreater \ x= \frac{-490}{49}=-10$

Answer 2

avem: x(1+3+5+...+13) = - 490
               13=2n-1  ⇒  n= (13+1):2=7 ...  termeni are suma din paranteza !
           x * ( 7^2 ) = - 490
                    x = -490:49 = -10 ∈ Z .