x(1+3+5+...+99)=2x(1+2+3+...50)-50
folosim formula generala pentru Suma lui Gauss:
1+2+3+...+n=[tex] \frac{n(n+1)}{2} [/tex]
respectiv pentru suma numerelor impare consecutive:
1+3+5+...+(2k-1)=[tex] k^{2} [/tex]
Deci exercitiul devine:
x[1+3+5+...+(2*50-1)]=2x*[tex] \frac{50*51}{2} [/tex] - 50
x*[tex] 50^{2} [/tex]=x*50*51 - 50
50=x*50(51-50)
50=x*50
x=1
