Question

Aratati ca numerele: a) 1+3+5+...+1999 b) 2020+2×(1+2+3+...+2019) sunt p.p

Answer 1

 

$\displaystyle\bf\\a)\\\\1+3+5+...+1999=?\\\\Calculam~numarul~de~termeni\!:\\\\n=\frac{1999-1}{2}+1=\frac{1998}{2}+1=999+1=\boxed{\bf1000~de~termeni}\\\\ Calculam~suma~cu~Gauss.\\\\1+3+5+...+1999=\frac{n(1999+1)}{2}=\\\\=\frac{1000\times2000}{2}=1000\times1000=\boxed{\bf1000^2=patrat~perfect}$

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$\displaystyle\bf\\b)\\\\2020+2\times(1+2+3+...+2019)=\\\\=2020+2\times\frac{2019(2019+1)}{2}=~~~~(Se~simplifica~2~cu~2)\\\\=2020+2019(2019+1)=\\\\=2020+2019\times2020=~~~~~Dam~factor~comun~pe~2020\\\\=2020(1+2019)=2020\times2020=\boxed{\bf2020^2~~~este~patrat~perfect}$