[tex] \displaystyle\\\text{Explicatii:}\\\text{Formula pentru o suma de puteri }\\ \text{cu aceeasi baza si exponenti consecutivi este:}\\\\k^0+k^1+k^2+k^3+\hdots + k^n=\frac{k^{n+1} -1}{k-1}\\\\\text{Aceasta formula este valabila doar daca exponentii incep de la 0.}\\\\\text{Daca primul termen }~k^0~\text{ lipseste, atunci folosim aceeasi formula}\\ \text{din care il scadem pe }~k^0~\text{care este egal cu 1}\\\\\text{Daca lipsesc mai multi ii scadem din formula.}\\\\ [/tex]
[tex] \displaystyle\\\text{In problema ta vom ajunge la:}\\\\3^1+3^2+3^3+\hdots+3^{2011}+3^{2012}=\frac{3^{2012+1}-1}{3-1}-3^0 =\\\\= \frac{3^{2013}-1}{2}-1 =\frac{3^{2013}-1}{2}-\frac{2}{2} =\frac{3^{2013}-1-2}{2}= \boxed{\bf \frac{3^{2013}-3}{2}}\\\\\text{Se observa ca l-am scazut pe }~3^0~\text{ care lipseste din sir.} [/tex]
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[tex] \displaystyle\\
\text{Rezolvare:}\\\\
\frac{3^{2013}}{5}-\frac{2\cdot3^{2012}}{5}- \frac{2\cdot3^{2011}}{5}-...- \frac{2\cdot3}{5}=\\\\
=\frac{3^{2013}}{5}-\frac{2}{5}\Big(3^{2012}+3^{2011}+...+3\Big)=\\\\
=\frac{3^{2013}}{5}-\frac{2}{5}\Big(3^1+3^2+3^3+...+3^{2011}+ 3^{2012} \Big)=\\\\
=\frac{3^{2013}}{5}-\frac{2}{5}\cdot\frac{3^{2013}-3}{2}=~~~\text{Conform formulei discutate mai sus.}\\\\
=\frac{3^{2013}}{5}-\frac{3^{2013}-3}{5}=\\\\
=\frac{3^{2013}-3^{2013}+3}{5}=\boxed{\bf \frac{3}{5}}
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