Đáp án:

      `n^ 2 - 4034.n + 4068289=1`

Giải thích các bước giải:

`1/8+1/24+1/48+.........+1/(n(n+2))=252/1009`

`=>1/2.4+1/4.6+1/6.8+.........+1/(n(n+2))=252/1009`

`=>2.(1/2.4+1/4.6+1/6.8+.........+1/(n(n+2)))=2. 252/1009`

`=>2/2.4+2/4.6+2/6.8+.........+2/(n(n+2))=504/1009`

`=>1/2-1/4+1/4-1/6+1/6-1/8+.........+1/n-1/(n+2)=504/1009`

`=>1/2-1/(n+2)=504/1009`

`=>1/(n+2)=1/2-504/1009`

`=>1/(n+2)=1/2-504/1009=1/2018`

`=>n+2=2018`

`=>n=2016`

Xét `n^ 2 - 4034.n + 4068289`

`=n^ 2 - 2.n.2017 + 2017^2`

`=(n-2017)^2`   (hàng đẳng thức)

Thay `n=2016` vào biểu thức, ta có:

  `(2016-2017)^2`

`=(-1)^2`

`=1`

  Vậy biểu thức có giá trị bằng `1.`

$\frac{1}{8}+$ $\frac{1}{24}+$ $\frac{1}{48}+...+$ $\frac{1}{n(n+2)}=$ $\frac{252}{1009}$

$⇒\frac{1}{2.4}+$ $\frac{1}{4.6}+...+$ $\frac{1}{n(n+2)}=$ $\frac{252}{1009}$

$⇒\frac{2}{2.4}+$ $\frac{2}{4.6}+...+$ $\frac{2}{n(n+2)}=$ $\frac{504}{1009}$ (nhân cả 2 vế với 2)

$⇒\frac{1}{2}-$ $\frac{1}{4}+$ $\frac{1}{4}-$ $\frac{1}{6}+...+$ $\frac{1}{n}-$ $\frac{1}{n+2}=$ $\frac{504}{1009}$

$⇒\frac{1}{2}-$ $\frac{1}{n+2}=$ $\frac{504}{1009}$

$⇒\frac{1}{n+2}=$ $\frac{1}{2018}$

$⇒n+2=2018$

$⇒n=2016$

$n^2-4034.n+4068289$

$=2016^2-4034.2016+4068289$

$=2016^2-4032.2017+(2017)^2$

$=2016^2-2017.(4032-2017)$

$=2016^2-2017.2015$

$=2016.(2017-1)-2017.2015$

$=2016.2017-2016-2017.2015$

$=2017.(2016-2015)-2016$

$=2017.1-2016$

$=1$