Observam ca termenii din prima paranteza sunt in progresie geometrica cu ratia q=-1/3.
[tex] S_{2012}=b_1*\frac{q^{2012}-1}{q-1}=1*\frac{(-{\frac{1}{3})}^{2012}-1}{\frac{1}{3}-1}=\frac{\frac{1}{3^{2012}}-1}{\frac{-2}{3}}=\frac{\frac{1-3^{2012}}{3^{2012}}}{\frac{-2}{3}}=\frac{1-3^{2012}}{3^{2012}}*\frac{3}{-2}=-\frac{1}{2}*\frac{1-3^{2012}}{3^{2011}} [/tex]
Sa vedem cum arata si a doua paranteza:
[tex] 1-\frac{1}{3^{2012}} =\frac{3^{2012}-1}{3^{2012}} [/tex]
Si acum sa vedem rezultatul final:
[tex] -\frac{1}{2}*\frac{1-3^{2012}}{3^{2011}}:\frac{3^{2012}-1}{3^{2012}}=-\frac{1}{2}*\frac{(-1)*(3^{2012}-1)}{3^{2011}}*\frac{3^{2012}}{3^{2012}-1}=-\frac{1}{2}*(-1)*3=\frac{3}{2} [/tex]
