Question

fractia ireductibila echivalenta cu fractia 11+22+33+...+ 110 supra 13+26+39+...+130 este a. 1 supra 3 b. 11 supra 13 c. 11 supra 3 d. 13 supra 11 ​

Answer 1

Răspuns:

11/13

Explicație pas cu pas:

Formula lui Gauss spune ca 1+2+3+4+5+6+...+n=nx(n+1):2

axb+axc+axd+...+axn=ax(b+c+d+...+n)

A=11+22+33+...+110=11x(1+2+3+...+10)=11x[10x(10+1):2]=11x(10x11:2)=11x(110:2)= 11x55=605

B=13+26+39+...+130=13x(1+2+3+...+10)=13x[10x(10+1):2]=13x(10x11:2)=13x(110:2)= 13x55=715

A/B=605/715=11X55/13X55=11/13