A = $\frac{1}{2^{2}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{4^{2}}$ +…+ $\frac{1}{9^{2}}$ Chứng minh $\frac{8}{9}$ > A > $\frac{2}{5}$

0 bình luận về “A = $\frac{1}{2^{2}}$ + $\frac{1}{3^{2}}$ + $\frac{1}{4^{2}}$ +…+ $\frac{1}{9^{2}}$ Chứng minh $\frac{8}{9}$ > A > $\frac{2}{5}$”

1. A = $\frac{1}{2^{2}}$ + $\frac{1}{3^{2}}$ + … + $\frac{1}{9^{2}}$ 

   Có:

   $\frac{1}{2^{2}}$ > $\frac{1}{2.3}$ 

   $\frac{1}{3^{2}}$ > $\frac{1}{3.4}$ 

   …

   $\frac{1}{9^{2}}$ > $\frac{1}{9.10}$ 

   A > $\frac{1}{2.3}$.$\frac{1}{3.4}$ …$\frac{1}{9.10}$ 

       = $\frac{1}{2}$ – $\frac{1}{10}$ 

       = $\frac{2}{5}$ 

   ⇒ A > $\frac{2}{5}$ (1)

   Lại có: 

   $\frac{1}{2^{2}}$ < $\frac{1}{1.2}$ 

   $\frac{1}{3^{2}}$ < $\frac{1}{2.3}$ 

   …

   $\frac{1}{9^{2}}$ < $\frac{1}{8.9}$

   A < $\frac{1}{1.2}$.$\frac{1}{2.3}$ …$\frac{1}{8.9}$

       = 1- $\frac{1}{9}$ 

       = $\frac{8}{9}$ 

   ⇒ A < $\frac{8}{9}$ (2)

   Từ 1 và 2 ⇒ $\frac{8}{9}$ > A > $\frac{2}{5}$

    

   Trả lời

2. Ta có : `A = (1/2^2 + 1/3^2 + 1/4^2 + …. + 1/9^2)`

   `-> (1/(2 . 2) + 1/(3 . 3) + 1/(4 . 4) + …. + 1/(9 . 9) < 1/(1 . 2) + 1/(2 . 3) + … + 1/(8 . 9)`

   Xét : `1/(1 . 2) + 1/(2 . 3) + …. + 1/(8 . 9)`

   `= 1 – 1/2 + 1/2 – 1/3 + …. + 1/8 – 1/9`

   `= 1 + ( – 1/2 + 1/2 – 1/3 + …. + 1/8) – 1/9`

   `= 1 – 1/9`

   `= 8/9`

   `-> A < 8/9 (1)`

   Ta có : `(1/(2 . 2) + 1/(3 . 3) + 1/(4 . 4) + …. + 1/(9 . 9) > 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/9 – 1/10`

   Xét `1/2 – 1/3 + 1/3 – 1/4 +…+ 1/9 – 1/10`

   `= 1/2 + (- 1/3 + 1/3 – 1/4 +…+ 1/9) – 1/10`

   `= 1/2-1/10`

   `= 2/5`

   `-> A > 2/5 (2)`

   Từ `(1), (2) -> đpcm`

   Trả lời