Question

Cho biểu thức M = \(\dfrac{1}{1!}+\dfrac{1}{2!}+......+\dfrac{1}{100!}\)

Chứng minh 3! - M > 4

Gíup mk vs............Mai mk ik học r

Answer 1

Ta có:

\(3!-M>4\\ 6-M>6-2\\ -M>-2\\ M< 2\)

Điều phải chứng minh: \(M< 2\)

\(M=\dfrac{1}{1!}+\dfrac{1}{2!}+...+\dfrac{1}{100!}\)

Ta có:

\(\dfrac{1}{2!}=\dfrac{1}{1\cdot2}\\ \dfrac{1}{3!}=\dfrac{1}{1\cdot2\cdot3}=\dfrac{1}{2\cdot3}\\ \dfrac{1}{4!}=\dfrac{1}{1\cdot2\cdot3\cdot4}< \dfrac{1}{3\cdot4}\\ \dfrac{1}{5!}=\dfrac{1}{1\cdot2\cdot3\cdot4\cdot5}< \dfrac{1}{4\cdot5}\\ ...\\ \dfrac{1}{100!}=\dfrac{1}{1\cdot2\cdot3\cdot...\cdot100}< \dfrac{1}{99\cdot100}\)

\(\Rightarrow M< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\\ M< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ M< 2-\dfrac{1}{100}< 2\)

Vậy \(M< 2\)

Vì

\(M< 2\\ \Rightarrow-M>-2\\ \Rightarrow6-M>6-2\\ \Leftrightarrow3!-M>4\left(đpcm\right)\)