\(C=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{18.19.20}\\ \Leftrightarrow C=1.\dfrac{1}{20}\\ C=\dfrac{1}{20}\)

theo mình thì

Ta 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 2C = 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

=> C = 189/ 760

mình nhầm

Ta có :

\(\dfrac{1}{1.2}-\dfrac{1}{2.3}=\dfrac{3}{1.2.3}-\dfrac{1}{1.2.3}=\dfrac{2}{1.2.3}\)

\(\dfrac{1}{2.3}-\dfrac{1}{3.4}=\dfrac{4}{2.3.4}-\dfrac{2}{2.3.4}=\dfrac{2}{2.3.4}\)

...

Do đó :

\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}\right)\)

\(\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{2.3}-\dfrac{1}{3.4}\right)\)

Vậy :

\(M=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{10.11}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{11.12}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{132}\right)\)

\(=\dfrac{1}{2}.\dfrac{65}{132}=\dfrac{65}{264}\)

sao chả ai trả lời cho tui thế

\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)

\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{19.20}< \dfrac{1}{1.2}\)

\(\Rightarrow2A< \dfrac{1}{2}\)

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

lỗi

lỗi cực kỳ

\(\dfrac{1}{2}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\) 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

a, A= 1/2. (2/1.2.3+2/2.3.4+2/3.4.5+...+2/18.19.20) A=1/2. (1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+...+1/18.19-1/19.20) A=1/2. (1/1.2-1/19.20) A=1/2. 189/380 A= 189/760

\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\)

\(A=\dfrac{1}{1}-\dfrac{1}{20}\)

\(A=\dfrac{20}{20}-\dfrac{1}{20}\)

\(A=\dfrac{19}{20}\)

A = \(\dfrac{1}{1.2.3}\)+\(\dfrac{1}{2.3.4}\)+\(\dfrac{1}{3.4.5}\)+...+\(\dfrac{1}{18.19.20}\)

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

A = \(\dfrac{1}{1.2}\)-\(\dfrac{1}{19.20}\)

A = \(\dfrac{1}{2}\)-\(\dfrac{1}{380}\)

A = \(\dfrac{189}{380}\)

(Mình nghĩ là vậy, có gì sai bạn bỏ qua nha )

a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)