So sánh tổng sau với ` 1`. `S = 1/3 + 1/7 + 1/13 + 1/21+…+ 1/91 + 1/111`

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1. Đáp án+Giải thích các bước giải:

   $\\S=\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{13}+…+\dfrac{1}{91}+\dfrac{1}{111}$ $\\\text{Ta có:}$$\\\dfrac{1}{3}<\dfrac{1}{2}\\\dfrac{1}{7}<\dfrac{1}{6}\\\dfrac{1}{13}<\dfrac{1}{12}\\………\\\dfrac{1}{91}<\dfrac{1}{90}\\\dfrac{1}{111}<\dfrac{1}{110}$ $\\=>\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{1}{13}+…+\dfrac{1}{91}+\dfrac{1}{111}<\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+…+\dfrac{1}{90}+\dfrac{1}{110}=\dfrac{1}{2.1}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+…+\dfrac{1}{9.10}+\dfrac{1}{10.11}\\=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+…+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\\=1-\dfrac{1}{11}\\=\dfrac{10}{11}<1\\=>S<1$

    

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2. Đặt A = 1/3 + 1/7 + 1/13 + 1/21 + … + 1/91 +1/111 

   Ta có:

   A < 1/2 + 1/6 + 1/12 + 1/20 + … +1/90 + 1/110

   <=> A < 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + … + 1/9.10 + 1/10.11

   <=> A < 1 – 1/2 + 1/2 – 1/3 +1/3 – 1/4 + 1/4 – 1/5 + … + 1/9 – 1/10 + 1/10 -1/11

   <=> A < 1 – 1/11 < 1 (đpcm) 

   Trả lời đầu vote 5* nhé

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