Chứng minh : $\frac{1}{3}$ + $\frac{1}{7}$ + $\frac{1}{13}$ + $\frac{1}{21}$ +$\frac{1}{31}$ + $\frac{1}{43}$ + $\frac{1}{57}$ + $\frac{1}{73}$ + $\

Chứng minh :
$\frac{1}{3}$ + $\frac{1}{7}$ + $\frac{1}{13}$ + $\frac{1}{21}$ +$\frac{1}{31}$ + $\frac{1}{43}$ + $\frac{1}{57}$ + $\frac{1}{73}$ + $\frac{1}{91}$ < 1

0 bình luận về “Chứng minh : $\frac{1}{3}$ + $\frac{1}{7}$ + $\frac{1}{13}$ + $\frac{1}{21}$ +$\frac{1}{31}$ + $\frac{1}{43}$ + $\frac{1}{57}$ + $\frac{1}{73}$ + $\”

  1. Đáp án:

    `A=1/3+1/7+1/13+1/21+1/31+1/43+1/57+1/73+1/91`

    `=>A=1/(1xx2+1)+1/(2xx3+1)+1/(3xx4+1)+1/(4xx5+1)+1/(5xx6+1)+1/(6xx7+1)+1/(7xx8+1)+1/(8xx9+1)+1/(9xx10+1`

    `=>A<1/(1xx2)+1/(2xx3)+1/(3xx4)+1/(4xx5)+1/(5xx6)+1/(6xx7)+1/(7xx8)+1/(8xx9)+1/(9xx10`

    `=>A<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10`

    `=>A<1-1/10<1`

    `=>A<1`

    Vậy biểu thức có giá trị bé hơn 1.

     

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  2. Gọi $\frac{1}{3}$ + $\frac{1}{7}$ + $\frac{1}{13}$ +…+ $\frac{1}{91}$ = A

    Ta có :

    $\frac{1}{3}$ < $\frac{1}{1.2}$

    $\frac{1}{7}$ < $\frac{1}{2.3}$

    $\frac{1}{13}$ < $\frac{1}{3.4}$

    ………….

    ………….

    $\frac{1}{91}$ < $\frac{1}{9.10}$

    ⇒A < $\frac{1}{1.2}$ + $\frac{1}{2.3}$ + $\frac{1}{3.4}$ + … + $\frac{1}{9.10}$

    ⇒A < 1 – $\frac{1}{2}$ + $\frac{1}{2}$ – $\frac{1}{3}$ + $\frac{1}{3}$ – $\frac{1}{4}$ + … + $\frac{1}{9}$ – $\frac{1}{10}$ 

    ⇒A < 1 – $\frac{1}{10}$ 

    ⇒A < $\frac{9}{10}$ 

    Mà $\frac{9}{10}$ < 1 ⇒ A < 1 

    ⇒ ĐPCM

     

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