Đáp án:
$a,$
`1/(1 . 2) + 1/(2 . 3) + 1/(3 . 4) + ... + 1/(19 . 20)`
`= 1 - 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/19 - 1/20`
`= 1 + (- 1/2 + 1/2 - 1/3 + 1/3 -1/4 + ... + 1/19) - 1/20`
`= 1 - 1/20`
`= 19/20`
$b,$
`B = 3/(n - 3)`
Để `B` nguyên
`-> 3 \vdots n - 3`
`-> n - 3 ∈ Ư (3) = {±1; ±3}`
Ta có bảng :
$\begin{array}{|c|c|c|c|c|c|c|}\hline n-3& 1 & -1 & 3 & -3 \\\hline n& 4 & 2 & 6& 0\\\hline\end{array}$
Vậy `n ∈ {4;2;6;0}` để `B` nguyên
