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21 tháng 7 2019

Ta có : 1/2 = 0,5

            2/3 = 0,666...

=> 1/2 + 2/3 + ... + 99/100 = 0,5 + 0,666...+3/4 + ... + 99/100

                                           = 1,1,6666... + 3/4 + ... +99/100 > 1

=> 1/2 + 2/3 + ... + 99/100 > 1

 \(=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\le1\)

\(=\frac{2-1}{2}+\frac{3-1}{3}+\frac{4-1}{4}+...+\frac{100-1}{100}\)

 \(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}\le1\)

\(\Rightarrow1-\frac{1}{100}\le1\)

5 tháng 10 2015

\(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)

\(\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+....+\frac{100-1}{100!}\)

\(1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+....+\frac{1}{99!}-\frac{1}{100!}\)

\(1-\frac{1}{100!}

14 tháng 11 2023

2:

\(B=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\cdot...\cdot\left(\dfrac{1}{100^2}-1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{100}+1\right)\)

\(=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\cdot...\cdot\left(\dfrac{1}{100}-1\right)\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{3}+1\right)\cdot...\cdot\left(\dfrac{1}{100}+1\right)\)

\(=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{101}{100}\)

\(=-\dfrac{1}{100}\cdot\dfrac{101}{2}=\dfrac{-101}{200}< -\dfrac{100}{200}=-\dfrac{1}{2}\)