Question

Arătați ca următorul nr e pătrat perfect:
a=1+3+5+...+2011

Answer 1

$\displaystyle 1+3+5+...+2011= \\ \\ =1+2+3+4+5+...+2011-(2+4+6+...+2010)= \\ \\ = \frac{2011(2011+1)}{2} -2(1+2+3+...+1005)= \\ \\ =\frac{2011 \times 2012}{2} -2 \times \frac{1005(1005+1)}{2} = \frac{4046132}{2} -2 \times \frac{1005 \times 1006}{2} = \\ \\ =2023066-\not 2 \times \frac{1011030}{\not2} =2023066-1011030= \\ \\ =1006(2011-1005)=1006 \times 1006=1006^2-p.p$