2.a)
[tex]\displaystyle \bf\\
2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+\cdots\\
\cdots+2^{2009}+2^{2010}+2^{2011}+2^{2012}\\\\
\text{Observam suma primilor 4 termeni este divizibila cu 15: }\\\\
2^1+2^2+2^3+2^4 = 2+4+8+16 = 30 ~\vdots ~15\\\\
\text{Vom grupa sirul in grupe de cate 4 termeni.}\\
\text{Avem voie sa facem asta deoarece avem 2012 termeni, iar }2012 ~\vdots~ 4.\\\\
\Big(2^1+2^2+2^3+2^4\Big)+\Big(2^5+2^6+2^7+2^8\Big)+\cdots\\
\cdots+\Big(2^{2009}+2^{2010}+2^{2011}+2^{2012}\Big)\\\\
[/tex]
[tex]\displaystyle \bf\\
\text{Din fiecare grupa dam factor comun:}\\\\
1\Big(2^1+2^2+2^3+2^4\Big)+2^4\Big(2^1+2^2+2^3+2^4\Big)+\cdots\\
\cdots+2^{2008}\Big(2^1+2^2+2^3+2^4\Big) =\\\\
=\Big(2^1+2^2+2^3+2^4\Big)\Big(1 + 2^4 + \cdots + 2^{2008}\Big)=\\\\
=30\Big(1 + 2^4 + \cdots + 2^{2008}\Big) ~\vdots ~15 [/tex]
2.b)
[tex]\displaystyle \bf\\
x^2 + y^2+4x-8y+20=0\\\\
\text{Descompunem pe 20 in suma de patrate perfecte.}\\
20 = 4 + 16\\\\
x^2 + y^2+4x-8y+4+16=0\\\\
(x^2 + 4x +4) + (y^2 -8y +16)=0\\\\
(x+2)^2(y-4)^2 = 0\\\\
\Longrightarrow~~~(x+2)^2 = 0~~~sau~~~(y-4)^2 = 0\\\\
x+2 = 0 \\\\
x = -2~~radacina ~dubla\\\\
y-4 = 0\\\\
y = 4 ~~radacina ~dubla
[/tex]
