Question

Ajutatima va rog !!!!

a=2009+2 x (1+2+3+...+2008)
b= 1+3+5+...+2009
Aratati ca e PATRAT PERFECT (p.p.)

Answer 1

$\displaystyle a)2009+2\cdot(1+2+3+...+2008)=2009+\not2 \cdot \frac{2008(2008+1)}{\not2} = \\ \\ =2009 +2008 \cdot 2009=2009(1+2008)=2009\cdot 2009=2009^2=p.p.$

$\displaystyle b).1+3+5+...+2009 \\ \\ 2009=1+(n-1) \cdot 2 \Rightarrow 2009=1+2n-2 \Rightarrow 2n=2010 \Rightarrow \\ \\ \Rightarrow n= \frac{2010}{2} \Rightarrow n=1005 \\ \\ S_{1005}= \frac{2 \cdot 1+(1005-1) \cdot 2}{2} \cdot 1005 \\ \\ S_{1005}= \frac{2+1004 \cdot 2}{2} \cdot 1005\\ \\ S_{1005}= \frac{2+2008}{2} \cdot 1005\\ \\ S_{1005}= \frac{2010}{2} \cdot 1005 \\ \\ S_{1005}=1005 \cdot 1005 \\ \\ S_{1005}=1005^2=p.p.$