Chứng minh : $\frac{1.2-1}{2!}$+$\frac{2.3-1}{3!}$+$\frac{3.4-1}{4!}$+$\frac{99.100-1}{100!}$<2

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1. Đáp án:

   Ta có : 

   `(1.2 – 1)/(2!) + (2.3 – 1)/(3!) + (3.4 – 1)/(4!) + …. + (99.100 – 1)/(100!)`

   `= (1.2)/(2!) – 1/(2!) + (2.3)/(3!) – 1/(3!) + (3.4)/(4!) – 1/(4!) + …. + (99.100)/(100!) – 1/(100!)`

   `= ((1.2)/(2!) + (2.3)/(3!) + (3.4)/(4!) + ….. + (99.100)/(100!) – (1/(2!) + 1/(3!) + 1/(4!) + …. + 1/(100!))`

   ` = (1 + 1 + 1/(2!) + …. + 1/(98!)) – (1/(2!) + 1/(3!) + 1/(4!) + ….. + 1/(100!))`

   ` = 2 – (1/(99!) + 1/(100!)) < 2`

   `=> đpcm`

   Giải thích các bước giải:

    

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2. Đặt $A = \dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+…+\dfrac{99.100-1}{100!}$

   $⇒ A =\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+…+\dfrac{99.100}{100!}-\dfrac{1}{100!}$

   $⇒ A=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+…+\dfrac{99.100}{100!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+…+\dfrac{1}{100!}\right)$

   $⇒ A =\left(1+1+\dfrac{1}{2!}+…+\dfrac{1}{98!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+…+\dfrac{1}{100!}\right)$

   $⇒ A=2-\dfrac{1}{99!}-\dfrac{1}{100!} < 2$

   $⇒ A < 2$

   $⇒ ĐPCM$

   Xin hay nhất !

    

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