Question

cum se rezolva acest exercitiu: 1+1/3+1/3^2+1/3^3...........+1/3^2003.Unde ^ inseamnaridicat la putere

Answer 1

$Notez~ \frac{1}{3} =a. \\ \\ S=1+a+a^2+a^3...+a^{2003}= \\ \\ ~~~= \frac{a^{2004}-1}{a-1}= \\ \\ ~~~= \frac{ (\frac{1}{3} )^{2004}-1}{ \frac{1}{3}-1 } = \\ \\ ~~~= \frac{ \frac{1}{3^{2004}}-1 }{ -\frac{2}{3} } = \\ \\ ~~~= (\frac{1}{3^{2004}}-1) \cdot (- \frac{3}{2})= \\ \\ ~~~= \frac{1-3^{2004}}{3^{2004}} \cdot (- \frac{3}{2})=$

$.~~~= \frac{3^{2004}-1}{3^{2003}} \cdot \frac{1}{2}= \\ \\ ~~~= \frac{1}{2} (3-\frac{1}{3^{2003}} )$

Suma se mai poate calcula intr-un fel. Am avut anul trecut, la olimpiada locala o suma asemanatoare.

$S= 1+\frac{1}{3}+ \frac{1}{3^2}+ \frac{1}{3^3} +... + \frac{1}{3^{2003}} = \\ \\ ~~~= \frac{3^{2003}+3^{2002}+3^{2001}+...+1}{3^{2003}} = \\ \\ ~~~= \frac{ \frac{3^{2004}-1}{3-1} }{3^{2003}} = \\ \\ ~~~= \frac{3^{2004}-1}{2} \cdot \frac{1}{3^{2003}}= \\ \\ ~~~= \frac{1}{2}(3- \frac{1}{3^{2003}} )$

Answer 2

Avem suma termenilor unei progresii geometrice de ratie q=1/3, primul termen b1=1 si numarul de termeni 2004.
Sn=b1·(q^n-1)/(q-1)
S=1·[(1/3)^2004-1]/(1/3-1)=-3/2·(1/3^2004-1)=-3/2[(1-3^2004)/3^2004=
=3/2·(3^2004-1)/3^2004=1/2·(3^2004-1)/3^2003