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

\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) Gio thi tu ma lam ko thích viết nữa mệt

A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\)

Theo công thức:

\(\dfrac{2m}{b.\left(b+m\right).\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+m.2\right)}\)Ta có:

2A=\(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+...+\dfrac{2}{18.19.20}\)

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

2A=\(\dfrac{1}{2}-\dfrac{1}{19.20}\)

A=\(\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right):2\)

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

A=\(\dfrac{1}{2}.\dfrac{19.20-2}{2.19.20}\)

A=\(\dfrac{19.20-2}{2.2.19.20}\) < \(\dfrac{19.20}{2.2.19.20}\) = \(\dfrac{1}{4}\)

\(\Rightarrow\) A<\(\dfrac{1}{4}\)

mik xin loi phan Ta có

\(\dfrac{2m}{b.\left(b+m\right)\left(b+2m\right)}=\dfrac{1}{b.\left(b+m\right)}-\dfrac{1}{\left(b+m\right).\left(b+2m\right)}\)Ta có blablabla

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​
 

khó à nha

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

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

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

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

2B=\(\frac{2}{1.2.3}\)+.....+\(\frac{2}{18.19.20}\)

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

2B=\(\frac{1}{1.2}\)-\(\frac{1}{19.20}\)

B=\(\frac{1}{1.2}\):2-\(\frac{1}{19.20}\):2

B=\(\frac{1}{1.2}\).\(\frac{1}{2}\)-\(\frac{1}{19.20}\).\(\frac{1}{2}\)

=\(\frac{1}{4}\)-\(\frac{1}{19.20.2}\)<\(\frac{1}{4}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

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

\(2B=\frac{1}{1.2}-\frac{1}{19.20}\)

\(B=\left(\frac{1}{2}-\frac{1}{19.20}\right):2\)

\(B=\frac{189}{760}\)

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