`text{Đáp án + Giải thích các bước giải}`
Đặt `A = 1/3 + 2/3^2 +...+ 100/3^100`
`3A = 1 + 2/3 +...+ 100/3^99`
`⇒3A - A = ( 1 + 2/3 +...+ 100/3^99 ) - ( 1/3 + 2/3^2 +...+ 100/3^100 )`
`⇒2A=1+2/3+...+100/3^99-1/3-2/3^2-...-100/3^100`
`⇒2A = 1 + ( 1/3 + 1/3^2 +...+ 1/3^99 ) - 100/3^100`
Đặt:
`B = 1/3 + 1/3^2 +...+ 1/3^99`
`⇒3B = 1 + 1/3 +...+ 1/3^98`
`⇒3B- B = ( 1 + 1/3 +...+ 1/3^98 ) - ( 1/3 + 1/3^2 +...+ 1/3^99 )`
`⇒2B=1-1/3^99`
`⇒2A = 1 + ( 1 - 1/3^99 ) - 100/3^100`
`⇒ 2A = 1 + 1 - 1/3^99 - 100/3^100`
`⇒ A = 1 - 1/3^99 - 100/3^100 < 1`
`⇒ A < 1`
Vậy `A < 1`
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