Question

demonstrati ca a= 3 la 1+ 3 la 2+ 3 la 3+ ...+3 la 60 este divizibil cu A) 4, B) 13

Answer 1

(3^1+3^2+3^3)+...+(3^58+3^59+3^60)=
=3^1(3^3:3^1+3^2:3^1+3^1:3^1)+...+3^58(3^60:3^58+3^59:3^58+3^58:3^58)
=3^1(3^2+3^1+3^0)+...+3^58(3^2+3^1+3^0)
=3^1*13+...+3^58*13
=13*(3^1+3^4+3^7+...+3^58) este divizibil cu 13

(3^1+3+2)+(3^3+3^4)+...+(3^59+3^60)=
3^1(3^2:3^1+3^1:3^1)+3^3(3^4:3^3+3^3:3^3)+...+3^59(3^60:3^59+3^59:3^59)=
=3^1(3^1+3^0)+3^3(3^1+3^0)+...+3^59(3^1+3^0)
=3^1(3+1)+3^3(3+1)+...+3^59(3+1)
=3^1*4+3^3*4+...+3^59*4
=4(3^1+3^3+3^5+...+3^59) este divizibil cu 4