chứng minh (1/5*6 ) +(1/7*8) +…+(1/49*50)<1/4

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1.   $\frac{1}{5.6}$ + $\frac{1}{6.7}$ + $\frac{1}{7.8}$ +…+ $\frac{1}{49.50}$

   = $\frac{1}{5}$ – $\frac{1}{6}$ + $\frac{1}{6}$ – $\frac{1}{7}$ + $\frac{1}{7}$ – $\frac{1}{8}$ +…+ $\frac{1}{49}$ – $\frac{1}{50}$  

   = $\frac{1}{5}$ – $\frac{1}{50}$ 

   = $\frac{9}{50}$< $\frac{1}{5}$ < $\frac{1}{4}$  

     => ĐPCM

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2. Ta có:

   `\qquad 1/{5.6}+1/{6.7}+1/{7.8}+…+1/{48.49}+1/{49.50}`

   `= 1/5-1/6+1/6-1/7+1/7-1/8+…+1/{48}-1/{49}+1/{49}-1/{50}`

   `=1/5+(1/6-1/6)+(1/7-1/7)+…+(1/{49}-1/{49})-1/{50}`

   `=1/5-1/{50}={10}/{50}-1/{50}`

   `=9/{50}<{10}/{50}`

   `<1/5<1/4`

   $\\$

   Ta lại có:

   `\qquad 1/{5.6}+1/{7.8}+…+1/{49.50}`

   `<1/{5.6}+1/{6.7}+1/{7.8}+…+1/{49.50}<1/4`

   Vậy `1/{5.6}+1/{7.8}+…+1/{49.50}<1/4`

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