Question

Aratati ca 1+3+5+...+1997 este patrat perfect

Answer 1

$1+3+5+...+(2n-1)=n ^{2}$
$2n-1=1997$
$2n=1997+1$
$2n=1998$
$n=1998:2$
$n=999$
$1+2+5+...+(2*999-1)=999 ^{2}$
$\sqrt{ 999^{2} } =999$