Question

chứng tỏ tổng sau nhỏ hơn 1:A=3/1x4+3/4x7+3/7x10+.....+3/40x43

Answer 1

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+.....+\dfrac{3}{40.43}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+.....+\dfrac{1}{40}-\dfrac{1}{43}\)

\(A=1-\dfrac{1}{43}\)

\(A< 1\left(đpcm\right)\)

Answer 2

\(A=3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{40}-\dfrac{1}{43}\right)\)

\(=3\left(1-\dfrac{1}{43}\right)=\dfrac{126}{43}>1\)

... sai đâu không nhỉ??