Đáp án:
`a, n = 2000`
`b, n ∈ { 0 ; 4 }`
Giải thích các bước giải:
`A, 1/3 + 1/6 + 1/10 + ... + 2/(n(n + 1)) = 1999/2001`
`⇒ 2/6 + 2/12 + 2/20 + ... + 2/(n(n + 1)) = 1999/2001`
`⇒ 2 . (1/6 + 1/12 + 1/20 + ... + 1/(n(n + 1))) = 1999/2001`
`⇒ 2 . (1/(2.3) + 1/(3.4) + 1/(4.5) + ... + 1/(n(n + 1))) = 1999/2001`
`⇒ 2 . (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/n - 1/(n + 1)) = 1999/2001`
`⇒ 2 . (1/2 - 1/(n + 1)) = 1999/2001`
`⇒ 1/2 - 1/(n + 1) = 1999/2001 : 2`
`⇒ 1/2 - 1/(n + 1) = 1999/2001 . 1/2`
`⇒ 1/2 - 1/(n + 1) = 1999/4002`
`⇒ 1/(n + 1) = 1/2 - 1999/4002`
`⇒ 1/(n + 1) = 2001/4002 - 1999/4002`
`⇒ 1/(n + 1) = 1/2001`
`⇒ n + 1 = 2001`
`⇒ n = 2001 - 1`
`⇒ n = 2000`
Vậy `n = 2000`
`B,` Để `A = (2n + 5)/(3n + 1) ∈ N`
`⇒ 2n + 5` $\vdots$ `3n + 1`
Mà $\left\{ \begin{array}{l}3(2n + 5) \vdots 3n + 1\\2(3n + 1) \vdots 3n + 1\end{array} \right.$
`⇒ 3(2n + 5) - 2(3n + 1)` $\vdots$ `3n + 1`
`⇒ 6n + 15 - 6n - 2` $\vdots$ `3n + 1`
`⇒ 13` $\vdots$ `3n + 1` `(n ∈ N)`
`⇒ 3n + 1 ∈ Ư (13) = { 1 ; 13 }`
`⇒ n ∈ { 0 ; 4 }`
Vậy `n ∈ { 0 ; 4 }`
