chứng minh $\frac{1.2-1}{2!}$ +$\frac{2.3-1}{3!}$ +$\frac{3.4-1}{4!}$ +….+$\frac{99.100-1}{100!}$ < 2

0 bình luận về “chứng minh $\frac{1.2-1}{2!}$ +$\frac{2.3-1}{3!}$ +$\frac{3.4-1}{4!}$ +….+$\frac{99.100-1}{100!}$ < 2”

1. Giải thích các bước giải:

   Ta có:

   $A=\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+…+\dfrac{99.100-1}{100!}$

   $\to A=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+….+\dfrac{99.100}{100!}-\dfrac{1}{100!}$

   $\to A=(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+….+\dfrac{99.100}{100!})-(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+….+\dfrac{1}{100!})$

   $\to A=(1+1+\dfrac1{2!}+…+\dfrac{1}{98!})-(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+….+\dfrac{1}{100!})$

   $\to A=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}$

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

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2. Ta có :

   A=1×2-1/2!+2×3-1/3!+3×4-1/4!+……..+99×100-1/100

   => A=1×2/2!-1/2!+2×3/3!-1/3!+3×4/4!-1/4!+…………+99×100/100!-1/100!

   => A=(1×2/2!+2×3/3!+3×4/4!+…+99×100/100!)-(1/2!+1/3!+1/4!+….+1/100!)

   => A=(1+1+1/2!+…..+1/98!)-(1/2!+1/3!+1/4!+…+1/100!)

   => A=1+1-1/99!-1/100!

   => A= 2-1/99!-1/100!

   => A<2

   vì 2=2 mà A= 2-1/99!-1/100! nên A<2

    

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