Đáp án: $A$
Giải thích các bước giải:
Ta có :
$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}.\dfrac{n+2-n}{n(n+1)(n+2)}=\dfrac{1}{2}.(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}$
Suy ra :
$ \dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+..+\dfrac{1}{n(n+1)(n+2)}$
$=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3})+\dfrac{1}{2}(\dfrac{1}{2.3}-\dfrac{1}{3.4})+..+\dfrac{1}{2}(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)})$
$=\dfrac{1}{2}(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)})$
$=\dfrac{1}{2}(\dfrac{1}{2}-\dfrac{1}{(n+1)(n+2)})$
$=\dfrac{1}{2}(\dfrac{1}{2}-\dfrac{1}{n^2+3n+2})$
$=\dfrac{1}{2}.\dfrac{n^2+3n+2-2}{2(n^2+3n+2)}$
$=\dfrac{n^2+3n}{4n^2+12n+16)}$
$\rightarrow T=(1+4)(3+12)=75$
$\rightarrow A$
