`B=1/3+1/(3^2)+1/(3^3)+....+1/(3^{99})`
`⇒3B=1+1/3+1/(3^2)+....+1/(3^{98})`
`⇒3B-B=(1+1/3+1/(3^2)+....+1/(3^{98}))-(1/3+1/(3^2)+1/(3^3)+....+1/(3^{99}))`
`⇒2B=1-1/(3^{99})`
`⇒2B=(3^{99}-1)/(3^{99})`
`⇒B=(3^{99}-1)/(2.3^{99})`
Vậy `B=(3^{99}-1)/(2.3^{99})`.