Question

Cho S=\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+......+\dfrac{3}{40.43}+\dfrac{3}{43.46}\).Hãy chứng tỏ S<1

Answer 1

\(S=\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{43\cdot46}\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{43}-\dfrac{1}{46}\)

\(S=1-\dfrac{1}{46}< 1\)

Answer 2

S= \(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{40\cdot43}+\dfrac{3}{43\cdot46}\)

S= \(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{42}-\dfrac{1}{46}\)

S= \(1-\dfrac{1}{46}\)

S= \(\dfrac{45}{46}\)

Mà \(\dfrac{45}{46}< 1\)

\(\Rightarrow S< 1\)

Vậy S < 1