Question

Chứng minh rằng:A=\(\frac{3}{1.4}\)+\(\frac{3}{2.6}\)+\(\frac{3}{3.8}\)+....+\(\frac{1}{2012.1342}\)<1;5

Answer 1

\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...............+\dfrac{1}{2012.1342}\)

\(A=\dfrac{3}{1.4}+\dfrac{3}{2.6}+\dfrac{3}{3.8}+...........................+\dfrac{3}{2012.4026}\)

\(A=\dfrac{6}{2.4}+\dfrac{6}{4.6}+\dfrac{6}{6.8}+..........................+\dfrac{6}{4024.4026}\)

\(A=3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...................+\dfrac{2}{4024.4026}\right)\)

\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{8}+....................+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)

\(A=3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)

\(A=3.\dfrac{1}{2}-3.\dfrac{1}{4026}\)

\(A=1,5-\dfrac{3}{4026}< 1,5\)

Answer 2

Ta có

A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{1}{2012.1342}\)

A = \(\dfrac{3}{1.4}\) + \(\dfrac{3}{2.6}\) + \(\dfrac{3}{3.8}\) + ... + \(\dfrac{3}{2012.4026}\)

A = \(\dfrac{6}{2.4}\) + \(\dfrac{6}{4.6}\) + \(\dfrac{6}{6.8}\) + ... + \(\dfrac{6}{4024.4026}\)

A = \(3\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+\dfrac{2}{6.8}+...+\dfrac{2}{4024.4026}\right)\)

A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{4024}-\dfrac{1}{4026}\right)\)

A = \(3\left(\dfrac{1}{2}-\dfrac{1}{4026}\right)\)

A = 3.\(\dfrac{1}{2}\) - 3.\(\dfrac{1}{4026}\)

A = 1,5 - \(3.\dfrac{1}{4026}\) < 1,5

=> A < 1,5

=> đpcm

 

Answer 3

Cô @Bùi Thị Vân hình như có gì đó nhầm lẫn!!