Bài 3 :
$B = (\dfrac{1}{2}) + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{98} + (\dfrac{1}{2})^{99}$
$\dfrac{1}{2}B = (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{99} + (\dfrac{1}{2})^{100}$
$\dfrac{1}{2}B - B = [ (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{99} + (\dfrac{1}{2})^{100}] - [(\dfrac{1}{2}) + (\dfrac{1}{2})^2+ (\dfrac{1}{2})^3 +....+ (\dfrac{1}{2})^{98} + (\dfrac{1}{2})^{99}]$
$-0,5B = (\dfrac{1}{2})^{100} - \dfrac{1}{2}$
$B = \dfrac{(\dfrac{1}{2})^{100} - \dfrac{1}{2}}{-0,5}$
$B =( \dfrac{1}{2})^{100} - \dfrac{1}{2}:{\dfrac{-1}{2}}$
$B =- \dfrac{1}{2})^{100} - \dfrac{1}{2}.-2$
Vì $(- \dfrac{1}{2})^{100} - \dfrac{1}{2}.2 <1$
⇒ $B<1$