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14 tháng 8 2017

4A=1+1/4+1/42+......+1/498

4A - A = ( 1+1/4+1/42+..........+1/498) - ( 1/4+1/42+1/43+.......+1/499)

3A= 1-1/499

A= 1/3 - 1/499 : 3

Mà 1/499 : 3 > 0 => 1/3 - 1/499 : 3 < 1/3

                          Hay A < 1/3

 

14 tháng 8 2017

a/ Rút gọn:

\(A=\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{99}}.\)

=> \(4A=1+\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}\)

=> \(4A=1+\left(\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+....+\frac{1}{4^{98}}+\frac{1}{4^{99}}\right)-\frac{1}{4^{99}}\)

<=> \(4A=1+A-\frac{1}{4^{99}}\)

=> \(3A=1-\frac{1}{4^{99}}\)

=> \(A=\frac{1}{3}-\frac{1}{3.4^{99}}\)

b/ Ta có: \(A=\frac{1}{3}-\frac{1}{3.4^{99}}< \frac{1}{3}\)

26 tháng 8 2016

A = 1/1×2 + 1/2×3 + 1/3×4 + .. + 1/99×100

A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100

A = 1 - 1/100 < 1

26 tháng 8 2016

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

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

\(A=1-\frac{1}{100}< 1\)

=>  ĐPCM

24 tháng 2 2017

ta có \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).......\left(\frac{1}{10}-1\right)\)

\(A=-\left(\frac{1}{2}.\frac{2}{3}.....\frac{9}{10}\right)\)

\(A=-\frac{1}{10}\)

vi\(-\frac{1}{10}>-\frac{1}{9}\)

do đó A>\(\frac{-1}{9}\)

24 tháng 2 2019

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{19}\right)\left(1-\frac{1}{20}\right)\)

\(A=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)...\left(\frac{19}{19}-\frac{1}{19}\right)\left(\frac{20}{20}-\frac{1}{20}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{18}{19}.\frac{19}{20}\)

\(A=\frac{1.2.3...18.19}{2.3.4...19.20}\)

\(A=\frac{1}{20}\Leftrightarrow A>\frac{1}{21}\)

24 tháng 2 2019

\(A=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)

\(A=\frac{1}{2}.\frac{2}{3}......\frac{19}{20}=\frac{1}{20}>\frac{1}{21}\)

\(\text{Vậy: A lớn hơn 1/21}\)

12 tháng 4 2016

Ta có: \(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{100^2}-1\right)\)

\(A=\frac{-3}{2^2}.\frac{-8}{3^2}.\frac{-15}{4^2}...\frac{-9900}{100^2}\)

\(A=\frac{\left(-1\right).3}{2^2}.\frac{\left(-2\right).4}{3^2}.\frac{\left(-3\right).5}{4^2}...\frac{\left(-99\right).101}{100^2}\)

\(A=\cdot\frac{\left(-1\right).\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}\)

\(A=\left(-\frac{1}{100}\right).\frac{101}{2}\)

\(A=-\frac{101}{200}\)