`#Sad`

`A=(1.2-1)/(2!)+(2.3-1)/(3!)+(3.4-1)/(4!)+...+(999.1000-1)/(1000!) < 2`

`=1.2-1/2!+2.3-1/3!+3.4-1/4!+...+999.1000-1/1000!`

`=1. 2/2!-1/2!+2. 3/3!-1/3!+3. 4/4!-1/4!+...+999. 1000/1000!-1/1000!`

`=(1. 2/2!+2. 3/3!+3. 4/4! +...+99. 1000/1000)`

`=(1/2!+1/3!+1/4!+...+1/1000)`

`=(1+1+1/2!+...+1/998!)-(1/2!+1/3!+1/4!+...+1/1000)`

`=1+1-1/999!-1/1000`

`=2-1/999!-1/1000!`

Mà`2-1999!-11000! < 2`

`=> A <2`

Đáp án:

Giải thích các bước giải:

`A` = `(1.2-1)/(1.2)+(2.3-1)/(1.2.3)+(3.4-1)/(1.2.3.4)+...+(999.1000-1)/(1.2.3.4× .... ×1000)` < 2

`A` = `(1.2-1)/(2!)+(2.3-1)/(3!)+(3.4-1)/(4!)+...+(999.1000-1)/(1000!)` < 2 

$=1.2-\dfrac{1}{2!}+2.3-\dfrac{1}{3!}+3.4-\dfrac{1}{4!}+...+999.1000-\dfrac{1}{1000!}$

$=1. \dfrac{2}{2!}-\dfrac{1}{2!}+2. \dfrac{3}{3!}-\dfrac{1}{3!}+3. \dfrac{4}{4!}-\dfrac{1}{4!}+...+999. \dfrac{1000}{1000!}-\dfrac{1}{1000!}$

$=(1. \dfrac{2}{2!} +2. \dfrac{3}{3!}+3. \dfrac{4}{4!}+...+99. \dfrac{1000}{1000})$

$=(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{1000})$

$=(1+1+\dfrac{1}{2!}+...+\dfrac{1}{998!})-(\dfrac{1}{2!}$ + $\dfrac{1}{3!}!+\dfrac{1}{4!}+...+\dfrac{1}{1000})$

$=1+1-\dfrac{1}{999!}-\dfrac{1}{1000}$ $=2-\dfrac{1}{999!}-\dfrac{1}{1000}!$ 

`mà` `2-1999!-11000! < 2`

`=>` `A < 2`

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