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[tex]Se\ considera\ sirul\ (a_n)_{n \geq 1},\ dephinit\ prin\ a_1=4, a_2=8, a_{n+2}=\frac{a_{n+1}}{a_n}\\
\vee n \in N^*. Calculati\ suma\ S=a_1+a_2+.....+a_{2004}[/tex]


Răspuns :

[tex]Se~observa~ca~sirul~este~periodic: \\ \\ a_1=4~;~a_2=8~;~a_3=2~;~a_4= \frac{1}{4}~;~ a_5=\frac{1}{8}~;~a_6= \frac{1}{2} ~;\\ \\ a_7=4~;~a_8=8~:~............ \\ \\ Perioada~are~lungimea~de~6. \\ \\ S=(a_1+...+a_6)+(a_7+...+a_{12})+...+(a_{1999}+...+a_{2004})= \\ \\ =334(4+8+2+ \frac{1}{4}+ \frac{1}{8}+ \frac{1}{2})=...~.[/tex]