Question

1/3+1/3 mũ 2+...+1/3 mũ 100

Answer 1

Đặt `A=1/3+1/(3^2)+...+1/(3^100)`

`3A=1+1/3+...+1/(3^99)`

`3A-A=(1+1/3+...+1/(3^99))-(1/3+1/(3^2)+...+1/(3^100))`

`2A=1-1/(3^100)`

`A=(1-1/(3^100))/2`

Answer 2

Đặt \(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)

\(3\cdot A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(3A-A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\right)\)

\(2A=1-\dfrac{1}{3^{100}}\)

\(\Rightarrow A=\dfrac{3^{100}-1}{3^{100}}:2=\dfrac{3^{100}-1}{2\cdot3^{100}}\)

#\(Toru\)