[tex]Notez~ \frac{1}{3} =a. \\ \\ S=1+a+a^2+a^3...+a^{2003}= \\ \\ ~~~= \frac{a^{2004}-1}{a-1}= \\ \\ ~~~= \frac{ (\frac{1}{3} )^{2004}-1}{ \frac{1}{3}-1 } = \\ \\ ~~~= \frac{ \frac{1}{3^{2004}}-1 }{ -\frac{2}{3} } = \\ \\ ~~~= (\frac{1}{3^{2004}}-1) \cdot (- \frac{3}{2})= \\ \\ ~~~= \frac{1-3^{2004}}{3^{2004}} \cdot (- \frac{3}{2})= [/tex]
[tex].~~~= \frac{3^{2004}-1}{3^{2003}} \cdot \frac{1}{2}= \\ \\ ~~~= \frac{1}{2} (3-\frac{1}{3^{2003}} )[/tex]
Suma se mai poate calcula intr-un fel. Am avut anul trecut, la olimpiada locala o suma asemanatoare.
[tex]S= 1+\frac{1}{3}+ \frac{1}{3^2}+ \frac{1}{3^3} +... + \frac{1}{3^{2003}} = \\ \\ ~~~= \frac{3^{2003}+3^{2002}+3^{2001}+...+1}{3^{2003}} = \\ \\ ~~~= \frac{ \frac{3^{2004}-1}{3-1} }{3^{2003}} = \\ \\ ~~~= \frac{3^{2004}-1}{2} \cdot \frac{1}{3^{2003}}= \\ \\ ~~~= \frac{1}{2}(3- \frac{1}{3^{2003}} ) [/tex]
