Toán 1/1!+1/2!+1/3!+…+1/2021! so sánh với P= 2 18/07/2021 By Margaret 1/1!+1/2!+1/3!+…+1/2021! so sánh với P= 2
Đáp án: Ta có : `1/(1!) + 1/(2!) + 1/(3!) + ……….. +1/(2021!)` `= 1 + 1/(1.2) + 1/(1.2.3) + ……… +1/(1.2.3…2021)< 1+ 1/(1.2) + 1/(2.3) + ……. + 1/(2020.2021)` `=1 + 1 – 1/2 + 1/2 – 1/3 +…………. +1/2020 – 1/2021=1+1 – 1/2021<1+1=2` =>`1/(1!)+1/(2!)+1/(3!) + …………….+1/(2021!)<2` `text{ @toanisthebest}` Trả lời
Đáp án: Đặt `A = 1/(1!) + 1/(2!) + 1/(3!) + … + 1/(2021!)` `⇔ A = 1 + 1/(1 × 2) + 1/(1 × 2 × 3) + … + 1/(1 × 2 × 3 × … ×2021)` Vì \(\left\{ \begin{array}{l}1 = 1\\ \dfrac{1}{1×2}=\dfrac{1}{1×2}\\ \dfrac{1}{1×2×3} < \dfrac{1}{2×3}\\……….\\ \dfrac{1}{1×2×3×…×2021} < \dfrac{1}{2020 × 2021}\end{array} \right.\) Nên `A < 1 + 1/(1 × 2) + 1/(2 × 3) + … + 1/(2020 × 2021)` `-> A < 1 + 1 – 1/2 + 1/2 – 1/3 + … + 1/2020 – 1/2021` `-> A < 1 + 1 + (- 1/2 + 1/2 – 1/3 + … + 1/2020) – 1/2021` `-> A < 2 – 1/2021` Ta thấy : `2 – 1/2021 < 2` `-> A < 2 – 1/2021 < 2` `-> A < 2` hay `A < P` Trả lời
Đáp án:
Ta có : `1/(1!) + 1/(2!) + 1/(3!) + ……….. +1/(2021!)`
`= 1 + 1/(1.2) + 1/(1.2.3) + ……… +1/(1.2.3…2021)< 1+ 1/(1.2) + 1/(2.3) + ……. + 1/(2020.2021)`
`=1 + 1 – 1/2 + 1/2 – 1/3 +…………. +1/2020 – 1/2021=1+1 – 1/2021<1+1=2`
=>`1/(1!)+1/(2!)+1/(3!) + …………….+1/(2021!)<2`
`text{ @toanisthebest}`
Đáp án:
Đặt `A = 1/(1!) + 1/(2!) + 1/(3!) + … + 1/(2021!)`
`⇔ A = 1 + 1/(1 × 2) + 1/(1 × 2 × 3) + … + 1/(1 × 2 × 3 × … ×2021)`
Vì \(\left\{ \begin{array}{l}1 = 1\\ \dfrac{1}{1×2}=\dfrac{1}{1×2}\\ \dfrac{1}{1×2×3} < \dfrac{1}{2×3}\\……….\\ \dfrac{1}{1×2×3×…×2021} < \dfrac{1}{2020 × 2021}\end{array} \right.\)
Nên `A < 1 + 1/(1 × 2) + 1/(2 × 3) + … + 1/(2020 × 2021)`
`-> A < 1 + 1 – 1/2 + 1/2 – 1/3 + … + 1/2020 – 1/2021`
`-> A < 1 + 1 + (- 1/2 + 1/2 – 1/3 + … + 1/2020) – 1/2021`
`-> A < 2 – 1/2021`
Ta thấy : `2 – 1/2021 < 2`
`-> A < 2 – 1/2021 < 2`
`-> A < 2`
hay `A < P`