Question

Aratati ca numarul natural N=1+3+5+7+.....+2005 este patrat perfect.

Answer 1

$\displaystyle 1+3+5+7+...+2005 \\ 2005=1+(n-1) \cdot 2 \\ 2005=1+2n-2 \\ 2n=2005-1+2 \\ 2n=2006 \\ n=2006:2 \\ n=1003 \\ S_{1003}= \frac{2+1002 \cdot 2}{2} \cdot 1003 \\ \\ S_{1003}= \frac{2+2004}{2} \cdot 1003 \\ \\ S_{1003}= \frac{2006}{2} \cdot 1003 \\ \\ S_{1003}=1003 \cdot 1003 \\ S_{1003}=1003^2-p.p$