Bài 5:
a)S= $3^{0}$+ $3^{2}$+ $3^{4}$+ $3^{6}$+ ...+ $3^{2002}$
S=1+ $3^{2}$+ $3^{4}$+ $3^{6}$+ ...+ $3^{2002}$
$3^{2}$ S= $3^{2}$+ $3^{4}$+ $3^{6}$+ ...+ $3^{2004}$
$3^{2}$ S-S=($3^{2}$+ $3^{4}$+ $3^{6}$+ ...+ $3^{2004}$)-(1+ $3^{2}$+ $3^{4}$+ $3^{6}$+ ...+ $3^{2002}$)
$3^{2}$ S-S= $3^{2004}$ - 1
Hay: S($3^{2}$ - 1)= $3^{2004}$ - 1
⇒8S= $x^{2004}$ - 1
⇒S= $\frac{ (3^{2004} -1)}{8}$
b) $\frac{1}{3}$- $\frac{2}{3^{2}}$+ $\frac{3}{3^{3}}$- $\frac{4}{3^{4}}$+ ...+ $\frac{99}{3^{99}}$- $\frac{100}{3^{100}}$
Gọi biểu thức trên là A, ta có:
3A= 1- $\frac{2}{3}$+ $\frac{3}{3^{2}}$- $\frac{4}{3^{3}}$+ ...+ $\frac{99}{3^{98}}$- $\frac{100}{3^{99}}$
⇒A+ 3A= ($\frac{1}{3}$- $\frac{2}{3^{2}}$+ $\frac{3}{3^{3}}$- $\frac{4}{3^{4}}$+ ...+ $\frac{99}{3^{99}}$- $\frac{100}{3^{100}}$) + (1- $\frac{2}{3}$+ $\frac{3}{3^{2}}$- $\frac{4}{3^{3}}$+ ...+ $\frac{99}{3^{98}}$- $\frac{100}{3^{99}}$) (1)
⇒4A . 3= 12A= 3 - 1+ $\frac{1}{3}$ - $\frac{1}{3^{2}}$ + ...+ $\frac{1}{3^{98}}$ - $\frac{1}{3^{99}}$ (2)
Cộng (1), (2) ta được:
16A= 3- $\frac{101}{3^{99}}$ - $\frac{100}{3^{100}}$
⇒ A= $\frac{3-\frac{101}{3^{99}} - \frac{100}{3^{100}} }{16}$
⇒ A= $\frac{3}{16}$ - $\frac{\frac{101}{3^{99}} - \frac{100}{3^{100}} }{16}$ <$\frac{3}{16}$
cho mik xin 5* và ctlhn
@chúc bn hok tốt
