Question

\(So.Sánh:\)

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

Answer 1

`A=1/3+1/[3^2]+1/[3^3]+...+1/[3^99]`

`3A=1+1/3+1/[3^2]+...+1/[3^98]`

`=>3A-A=1+1/3+1/[3^2]+...+1/[3^98]-1/3-1/[3^2]-1/[3^3]-...-1/[3^99]`

`=>2A=1-1/[3^99]`

`=>A=1/2-1/[2.3^99]`

 Vì `1/2 > 1/2-1/[2.3^99]`

`=>B > A`