Lời giải:
Gọi tổng trên là $A$

$A=2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200})$

$=2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{200-199}{199.200})$

$=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{199}-\frac{1}{200})$

$=2(\frac{1}{2}-\frac{1}{200})=1-\frac{1}{100}=\frac{99}{100}$

Đặt A = 1 x 2 + 2 x 3 + ... + 199 x 200

3A = 1 x 2 x 3 + 2 x 3 x (4-1) + .... + 199 x 200 x (201 - 198)

3A = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 +.... + 199 x 200 x 201 - 198 x 199 x 200

3A = ( 1 x 2 x 3 - 1 x 2 x 3) + ( 2 x 3 x 4 - 2 x 3 x 4) + ....... + (198 x 199 x 200 - 198 x 199 x 200) + 199 x 200 x 201

Do đó A = 67 x 200 x 199 = 2666600 

Đặt A=1x2+2x3+3x4+4x5+........+199x200

Ta có:

 3A=1x2x3+2x3x3+3x4x3+.......+199x200x3

3A=1x2x3+2x3x(4-1)+3x4x(5-2)+....+199x200x(201-198)

3A=1x2x3+2x3x4-1x2x3+3x4x5-2x3x4+.............+199x200x201-198x199x200

3A=199x200x201

A=39800x201:3

A=39800x67

A=2666600

Vậy 1x2+2x3+3x4+........+199x200=2666600

mấy bạn thấy chưa

Uchiha Nguyễnđã được 6 cái lik e của mình đó

Ta có: \(C=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+\dfrac{2}{5.6}+\dfrac{2}{6.7}\)

\(\Leftrightarrow C=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)

\(\Leftrightarrow C=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(\Leftrightarrow C=2\left(1-\dfrac{1}{7}\right)=\dfrac{2.6}{7}=\dfrac{12}{7}\)

A=2(\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\))=2(\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\))

=> A=2(\(\frac{1}{1}-\frac{1}{100}\))=2.\(\frac{99}{100}=\frac{99}{50}\)

ĐS: A=99/50

\(\frac{2}{1\times2}+\frac{2}{2\times3}+\frac{2}{3\times4}+\frac{2}{4\times5}+...+\frac{2}{99\times100}\)

\(=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+...+\frac{1}{99\times100}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}-\frac{1}{100}\)

\(=\frac{99}{100}\)

\(2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\times\left(1-\frac{1}{100}\right)\)

\(=2\times\frac{99}{100}\)

\(=\frac{99}{50}\)

Đặt \(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{99.100}\)

\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)\)

\(A=2.\left(1-\frac{1}{100}\right)\)

\(A=\frac{2.99}{100}\)

\(A=\frac{99}{50}=1\frac{49}{50}\)

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{99.100}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(=2\left(1-\frac{1}{100}\right)=2.\frac{99}{100}\)

\(=\frac{99}{50}\)

Ta thấy: \(\frac{1}{1.2}=\frac{2-1}{1.2}=\frac{2}{1.2}-\frac{1}{1.2}=1-\frac{1}{2}\); \(\frac{1}{2.3}=\frac{3-2}{2.3}=\frac{3}{2.3}-\frac{2}{2.3}=\frac{1}{2}-\frac{1}{3}\)

Tương tự với các phân số khác

Cho A=2/1x2 + 2/2x3 + 2/3x4 + 2/4x5 + ... + 2/19x20

=> \(A=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{18.19}+\frac{1}{19.20}\right)=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}\right)\)

\(A=2\left(1-\frac{1}{20}\right)=2.\frac{19}{20}=\frac{19}{10}=1,9\)

Chú ý dấu chấm là dấu nhân

\(\frac{2}{1\times2}+\frac{2}{2\times3}+...+\frac{2}{19\times20}\)

\(=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}\right)\)

\(=2\times\left(1-\frac{1}{20}\right)=2\times\frac{19}{20}=\frac{19}{10}\)

\(P=\left(1-\frac{2}{2.3}\right).\left(1-\frac{2}{3.4}\right).\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)

\(P=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)

\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)

\(P=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5...100}\)

\(P=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)