Đặt `A=3+3/[2]+3/[2^2]+3/[2^3]+...+3/[2^100]`
`=> A=3(1+1/2+1/[2^2]+1/[2^3]+...+1/[2^100])`
Đặt `B=1+1/2+1/[2^2]+1/[2^3]+...+1/[2^100]`
`=> 1/2B=1/2+1/[2^2]+1/[2^3]+1/[2^4]+...+1/[2^101]`
`=> B-1/2B=1+1/2+1/[2^2]+1/[2^3]+...+1/[2^100]-1/2-1/[2^2]-1/[2^3]-1/[2^4]-...-1/[2^101]`
`=> 1/2B=1-1/[2^101]`
$⇒ B=\dfrac{1-\dfrac{1}{2^{101}}}{\dfrac{1}{2}}$
$⇒ B=\dfrac{{2}}{1-\dfrac{1}{2^{101}}}$
$⇒ A=3.\dfrac{{2}}{1-\dfrac{1}{2^{101}}}$
$⇒ A=\dfrac{{6}}{1-\dfrac{1}{2^{101}}}$
