Đặt A = `1/3+2/3^2+3/3^3 +...+100/3^100`
`⇒3A= 1+2/3+3/3^2 +...+100/3^99`
`⇒3A-A=2A=1+1/3+1/3^2+....+1/3^99-1/3^100`
`⇒A=(1+1/3+1/3^2+....+1/3^99-1/3^100)/2`
Đặt `1+1/3+1/3^2+....+1/3^99=B`
`⇒3B= 3+1+1/3+....+1/3^98`
`⇒3B-B=2B=3-1/3^99`
`⇒B=(3-1/3^99)/2`
`⇒A=((3-1/3^99)/2-1/3^100)/2`