Đáp án:
Ta có :
`B = (3/2 - 2/2^2)(4/3 - 2/3^2)(5/4- 2/4^2) ..... (101/100 - 2/100^2)`
`= (6/2^2 - 2/2^2)(12/3^2 - 2/3^2)(20/4^2 - 2/4^2)....((101.100)/(100^2) - 2/100^2)`
`= 4/2^2 . 10/3^2 . 18/4^2 ..... ((101.100 - 2)/100^2)`
`= (1.4)/(2.2) . (2.5)/(3.3) . (3.6)/(4.4) ..... (99.102)/(100.100)`
`= (1.2.3...99)/(2.3.4...100) . (4.5.6...102)/(2.3.4....100)`
`= (1.2.3...99)/[(1.2.3...99).100] . [(4.5.6....100).101.102]/[2.3.(4.5.6....100)]`
`= 1/100 . (101.102)/(2.3)`
`= 1717/100`
Giải thích các bước giải:
