Question

Aratati ca numarul 1+2× [E(0)+E(1)+...+E(18)] este patrat perfect, unde E(x)= 2x+3/2
VA ROGGG​

Answer 1

$\displaystyle\bf\\E(0)+E(1)+...+E(n)=\frac{2\cdot 0+3}{2} + \frac{2\cdot 1 +3}{2} +...+ \frac{2n+3}{2} =\\\sum_{k=0}^n \frac{2k+3}{2} = \sum_{k=0}^n\frac{2k}{2} +\frac{3}{2} =\sum_{k=0}^n k+\sum_{k=0}^n\frac{3}{2} =\frac{1}{2} n(n+1)+\frac{3}{2}(n+1) =$                                    

$\displaystyle\bf\\\frac{1}{2}(n+1)(n+3),~deci~pentru~n=18~\implies E(0)+E(1)+...+E(18)=\\\frac{19\cdot 21}{2}.\\1+2\cdot(E(0)+E(1)+...+E(18))=1+2\cdot \frac{19 \cdot 21}{2} = 1+19\cdot21=400=20^2~\\care~este~patrat~perfect.\\desigur,~se~mai~putea~face~si~astfel~:\\\frac{2\cdot 0+3}{2} + \frac{2\cdot 1 +3}{2} +...+ \frac{2\cdot 18+3}{2}=$

$\displaystyle\bf\\\frac{2\cdot 0 +3 + 2\cdot 1 +3+...+2\cdot18+3}{2} = \frac{2(0+1+2+...+18)+3\cdot 19}{2} =\\\frac{2\cdot\frac{18\cdot 19}{2}+57 }{2} = \frac{342+57}{2} =\frac{399}{2}.\\inlocuim~:~ 1+2\cdot \frac{399}{2} =400=20^2=patrat~perfect.$