Question

Sa se arate ca numarul P=1+3+5+...+1997=? este patrat perfect

Answer 1

[tex]p=1+3+5+...+1997\\ p=1+2+3+...+1997-2-4-6-....-1996\\ p=\frac{1997*1998}{2}-2(1+2+3+....+998)\\ p=1997*999-\not2*\frac{998*999}{\not2}\\ p=1997*999-998*999\\ p=999(1997-998)\\ p=999^2[/tex]

Answer 2

$\displaystyle 1+3+5+...+1997 \\ 1997=1+(n-1) \cdot 2 \\ 1997=1+2n-2 \\ 2n=1997-1+2 \\ 2n=1998 \\ n=1998:2 \\ n=999 \\ S_{999}= \frac{2+998 \cdot 2}{2} \cdot 999 \\ \\ S_{999}= \frac{2+1996}{2} \cdot 999 \\ \\ S_{999}= \frac{1998}{2} \cdot 999 \\ \\ S_{999}=999 \cdot 999 \\ S_{999}=999^2-p.p$