Question

[Nguyễn Văn Kiên]

1.Cho A=\(\dfrac{1}{2^2}+\)\(\dfrac{1}{4^2}+\dfrac{1}{6^2}+....+\dfrac{1}{100^2}.\)Chứng minh A<\(\dfrac{1}{2}\)

2.Cho B=\(\dfrac{1}{3}+\dfrac{2}{9}+\dfrac{3}{27}+...+\dfrac{2019}{3^{2019}}\).Chứng minh:B<\(\dfrac{3}{4}\)

3.Rút gon biểu thức C=\(\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+1\dfrac{3}{5}\right).....\left(1+\dfrac{1}{2014.2016}\right)\)   giúp mik với

Answer 1

[Nguyễn Lê Phước Thịnh]

3: \(C=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2014\cdot2016}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2015^2-1}\right)\)

\(=\dfrac{2^2-1+1}{2^2-1}\cdot\dfrac{3^2-1+1}{3^2-1}\cdot...\cdot\dfrac{2015^2-1+1}{2015^2-1}\)

\(=\dfrac{2^2}{\left(2-1\right)\left(2+1\right)}\cdot\dfrac{3^2}{\left(3-1\right)\left(3+1\right)}\cdot...\cdot\dfrac{2015^2}{\left(2015-1\right)\left(2015+1\right)}\)

\(=\dfrac{2\cdot3\cdot...\cdot2015}{1\cdot2\cdot...\cdot2014}\cdot\dfrac{2\cdot3\cdot...\cdot2015}{3\cdot4\cdot...\cdot2016}\)

\(=\dfrac{2015}{1}\cdot\dfrac{2}{2016}=\dfrac{2015}{1008}\)

1: \(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\)

\(=\dfrac{1}{2^2}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)\)

\(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\)

...

\(\dfrac{1}{50^2}< \dfrac{1}{49\cdot50}=\dfrac{1}{49}-\dfrac{1}{50}\)

Do đó: \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

=>\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}< 1-\dfrac{1}{50}\)

=>\(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 2-\dfrac{1}{50}\)

=>\(A=\dfrac{1}{4}\left(1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\right)< \dfrac{1}{4}\left(2-\dfrac{1}{50}\right)\)

=>\(A< \dfrac{1}{2}-\dfrac{1}{200}\)

=>A<1/2