Question

Se considera numerele a=1+2+3+...+48 si b=1+2+3+...+63. Determinati raportul numerelor naturale a si b.

Answer 1

a=1+2+3+...+48=48•49/2=1176; b=1+2+3+...+63=63•64/2=2016; a/b=1176/2016=0,58(3)

Answer 2

a=1+2+3+...+48

 b=1+2+3+...+63.

sunt 2 sume gauss cu formula de calcul [n(n+1)]:2
$a= \frac{n(n+1)}{2}= \frac{48*49}{2} =1176 \\ \\ b=\frac{n(n+1)}{2}=\frac{63(63+1)}{2}=2016 \\ \\ \frac{a}{b}= \frac{1176}{2016} = \frac{588}{1008} = \frac{294}{505}=0,58(3)$