Đáp án:
Bài `3`
`1/3 + 1/6 + 1/10 + ... + 2/(n (n + 1) ) = 2005/2006`
`⇔ 2/6 + 2/12 + 2/20 + ... + 2/(n (n + 1) ) = 2005/2006`
`⇔ 2 [1/6 + 1/12 + 1/20 + ... + 1/(n (n + 1) ) ] = 2005/2006`
`⇔ 1/(2 × 3) + 1/(3 × 4) + 1/(4 × 5) + ... + 1/(n (n + 1) ) = 2005/2006 ÷ 2`
`⇔ 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/n - 1/(n +1) = 2005/4012`
`⇔ 1/2 + (- 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/n) - 1/(n + 1) = 2005/4012`
`⇔ 1/2 - 1/(n + 1) = 2005/4012`
`⇔ 1/(n + 1) = 1/2 - 2005/4012`
`⇔ 1/(n + 1) = 1/4012`
`⇔ 4012 = n + 1`
`⇔ n = 4012 - 1`
`⇔ n = 4011 (tm)`
Vậy `n= 4011`
Bài `4`
`1/(5 × 8) + 1/(8 × 11) + 1/(11 × 14) + ... + 1/(x ( x + 3) ) = 101/1540`
Nhân `VP,VT` với `3` ta đươc :
`⇔ 3 × [1/(5 × 8) + 1/(8 × 11) + 1/(11 × 14) + ... + 1/(x ( x + 3) )] = 3 × 101/1540`
`⇔ 3/(5 × 8) + 3/(8 × 11) + 3/(11 × 14) + ... + 3/(x ( x + 3) ) = 303/1540`
`⇔ 1/5 - 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + ... + 1/x - 1/(x + 3) = 303/1540`
`⇔ 1/5 + (- 1/8 + 1/8 - 1/11 + 1/11 - 1/14 + ... + 1/x) - 1/(x + 3) = 303/1540`
`⇔ 1/5 - 1/(x + 3) = 303/1540`
`⇔ 1/(x + 3) = 1/5 - 303/1540`
`⇔ 1/(x + 3) = 1/308`
`⇔ 308 = x + 3`
`⇔ x = 308 - 3`
`⇔ x = 305 (tm)`
Vậy `x = 305`
