[tex]\displaystyle\bf\\a)\\2+2\cdot3+2\cdot3^2+2\cdot3^3+...+2\cdot3^{2011}=x\cdot81^{502}-1\\\\\text{Dam factor comun pe 2.}\\\\2\Big(1+3+3^2+3^3+...+3^{2011}\Big)=x\cdot81^{502}-1\\\\2\Big(3^0+3^1+3^2+3^3+...+3^{2011}\Big)=x\cdot81^{502}-1\\\\Aplicam~formula:~~~k^0+k^1+k^2+...+k^n=\frac{k^{n+1}-1}{k-1}\\\\\\2\Big(3^0+3^1+3^2+3^3+...+3^{2011}\Big)=\\\\=2\cdot\frac{3^{2011+1}-1}{3-1}=2\cdot\frac{3^{2012}-1}{2}=\boxed{\bf3^{2012}-1}[/tex]
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[tex]\displaystyle\bf\\Rezolvam~ecuatia:\\\\3^{2012}-1=x\cdot81^{502}-1~~~\Big|+1\\\\3^{2012}=x\cdot81^{502}\\\\3^{2012}=x\cdot\Big(3^4\Big)^{502}\\\\3^{2012}=x\cdot3^{4\times502}\\\\3^{2012}=x\cdot3^{2008}\\\\x=\frac{3^{2012}}{3^{2008}}\\\\x=3^{2012-2008}\\\\\boxed{\bf~x=3^4}\\\\\boxed{\bf~x=81}[/tex]
