cho A=1/2-1/3+1/4-1/5+...+1/48-1/49 chứng minh 1/5<A<2/5

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Giải thích các bước giải:

$1-S-\dfrac1{50}=1-\dfrac 12+\dfrac 13-\dfrac 14+\dfrac 15-...-\dfrac 1{48}+\dfrac 1{49}-\dfrac1{50}$$\dfrac{49}{50}-S=1+\dfrac 12+\dfrac 13+\dfrac 14+\dfrac 15+...+\dfrac 1{48}+\dfrac 1{49}+\dfrac1{50}-2\left(\dfrac12+\dfrac14+\dfrac16+...+\dfrac1{50}\right)$$\dfrac{49}{50}-S=1+\dfrac 12+\dfrac 13+\dfrac 14+\dfrac 15+...+\dfrac 1{48}+\dfrac 1{49}+\dfrac1{50}-1-\dfrac12-\dfrac13-...-\dfrac1{25}$$\dfrac{49}{50}-S=\dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}$$S=\dfrac{49}{50}-\left(\dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}\right)$Ta có:$\dfrac1{26}>\dfrac1{30}\\\dfrac{1}{27}>\dfrac1{30}\\...\\\dfrac1{29}>\dfrac1{30}\\\dfrac1{31}>\dfrac1{40}\\\dfrac1{32}>\dfrac1{40}\\...\\\dfrac1{39}>\dfrac1{40}\\\dfrac1{41}>\dfrac1{50}\\\dfrac1{42}>\dfrac1{50}\\...\\\dfrac1{49}>\dfrac1{50}$$\Rightarrow \dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}>5.\dfrac1{30}+10.\dfrac1{40}+10.\dfrac1{50}=\dfrac16+\dfrac14+\dfrac15=\dfrac{37}{60}>\dfrac{36}{60}=\dfrac 35$$\Rightarrow S=\dfrac{49}{50}-\left(\dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}\right)<\dfrac{49}{50}-\dfrac 35 =\dfrac{19}{50}<\dfrac{20}{50}=\dfrac 25$Lại có:$\dfrac1{26}<\dfrac1{25}\\\dfrac{1}{27}<\dfrac1{25}\\...\\\dfrac1{30}<\dfrac1{25}\\\dfrac1{31}<\dfrac1{30}\\\dfrac1{32}<\dfrac1{30}\\...\\\dfrac1{40}<\dfrac1{30}\\\dfrac1{41}<\dfrac1{40}\\\dfrac1{42}<\dfrac1{40}\\...\\\dfrac1{50}<\dfrac1{40}$$\Rightarrow \dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}<5.\dfrac1{25}+10.\dfrac1{30}+10.\dfrac1{40}=\dfrac13+\dfrac14+\dfrac15=\dfrac{47}{60}<\dfrac{48}{60}=\dfrac 45$$\Rightarrow S=\dfrac{49}{50}-\left(\dfrac1{26}+\dfrac1{27}+\dfrac1{28}+...+\dfrac1{50}\right)>\dfrac{49}{50}-\dfrac {4}{5} =\dfrac{9}{50}>\dfrac{10}{50}=\dfrac {1}{5}$

(Câu trả lời hay nhất đi, vì tại sao không :) )