khó à nha

a chứng minh được bài toán tổng quát sau 

2/[(n-1)n(n+1)] = 1/[(n-1)n] - 1/[n(n+1)] 

Áp dụng: 

ta có 2A = 1/(1.2) - 1/ (2.3) +1/(2.3) - 1/(3.4) + ...+ 1/18.19 - 1/19.20 

= 1/(1.2) - 1/(19.20) = [190 - 1] / 19.20 = 189/380 

=> A = 189/ 760 < 1/4

=(1/1-1/2-1/3)+(1/2-1/3-1/4)+...+(1/18-1/19-1/20)

=1/1-1/20=19/20

k nha

dung luon nhung van tat qua

Lời giải:

Gọi tổng trên là $A$

$A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}$

$2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{18.19.20}$

$=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{20-18}{18.19.20}$

$=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{18.19}-\frac{1}{19.20}$

$=\frac{1}{1.2}-\frac{1}{19.20}=\frac{189}{380}$

$\Rightarrow A=\frac{189}{760}$

ĐÚNG ĐÓ BẠN ƠI

A = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20 

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(A=\frac{1}{4}-\frac{1}{2.19.20}< \frac{1}{4}\)

Nhận thấy: \(\dfrac{1}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{2}{2\cdot n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{2+n-n}{2n\cdot\left(n+1\right)\cdot\left(n+2\right)}\\ =\dfrac{1}{2}\cdot\left[\dfrac{2+n-n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{2+n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}-\dfrac{n}{n\cdot\left(n+1\right)\cdot\left(n+2\right)}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{n\cdot\left(n+1\right)}-\dfrac{1}{\left(n+1\right)\cdot\left(n+2\right)}\right]\)

\(\Rightarrow A=\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{18\cdot19\cdot20}\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\right]\\ =\dfrac{1}{2}\cdot\left[\dfrac{1}{1\cdot2}-\dfrac{1}{19\cdot20}\right]\\ =\dfrac{1}{4}-\dfrac{1}{760}< \dfrac{1}{4}\)

Vậy \(A< \dfrac{1}{4}\)

quá đỉnh:)

Nhận thấy: 1�⋅(�+1)⋅(�+2)=22⋅�⋅(�+1)⋅(�+2)=2+�−�2�⋅(�+1)⋅(�+2)=12⋅[2+�−��⋅(�+1)⋅(�+2)]=12⋅[2+��⋅(�+1)⋅(�+2)−��⋅(�+1)⋅(�+2)]=12⋅[1�⋅(�+1)−1(�+1)⋅(�+2)]n⋅(n+1)⋅(n+2)1​=2⋅n⋅(n+1)⋅(n+2)2​=2n⋅(n+1)⋅(n+2)2+n−n​=21​⋅[n⋅(n+1)⋅(n+2)2+n−n​]=21​⋅[n⋅(n+1)⋅(n+2)2+n​−n⋅(n+1)⋅(n+2)n​]=21​⋅[n⋅(n+1)1​−(n+1)⋅(n+2)1​]

⇒�=11⋅2⋅3+12⋅3⋅4+...+118⋅19⋅20=12⋅[11⋅2−12⋅3+12⋅3−13⋅4+...+118⋅19−119⋅20]=12⋅[11⋅2−119⋅20]=14−1760<14⇒A=1⋅2⋅31​+2⋅3⋅41​+...+18⋅19⋅201​=21​⋅[1⋅21​−2⋅31​+2⋅31​−3⋅41​+...+18⋅191​−19⋅201​]=21​⋅[1⋅21​−19⋅201​]=41​−7601​<41​

Vậy �<14A<41​
 

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