giúp mik nka gấp lắm

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phamhuy20011801

Đặt $A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}$
Thì $3A=\dfrac{3}{3}+\dfrac{3}{3^2}+\dfrac{3}{3^3}+...+\dfrac{3}{3^{100}}\\
=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}$
$\rightarrow 3A-A=2A=(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}})-(\dfrac{1}{3}+\dfrac{1}{3^2}+ \dfrac{1}{3^3}+...+\dfrac{1}{3^{100}})\\
=1-\dfrac{1}{3^{100}}\\
=\dfrac{3^{100}-1}{3^{100}}$

Suy ra $A=\dfrac{\dfrac{3^{100}-1}{3^{100}}}{2}=\dfrac{3^{100}-1}{2.3^{100}}$

 
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