Gọi biểu thức trên là A, ta có :

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

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 199x100x3

A x 3 = 1x2x3 + 2x3x(4-1) + 3x4x(5-2) + 4x5x(6-3) + ... + 99x100x(101-98)

A x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 199x200x101 - 198x199x100.

A x 3 = 199x200x101

A = 199x200x201:3

A=2666600

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

\(D=\dfrac{5}{1\cdot2}+...+\dfrac{5}{199\cdot200}\)

\(=\dfrac{5}{2}\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)

\(=\dfrac{5}{2}\cdot\dfrac{199}{200}=\dfrac{199}{80}\)

Lời giải:

\(D=5\times \left(\frac{1}{1\times 2}+\frac{1}{2\times 3}+\frac{1}{3\times 4}+...+\frac{1}{199\times 200}\right)\)

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

\(=5\times \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{199}-\frac{1}{200}\right)=5\times (1-\frac{1}{200})\)

\(=5\times \frac{199}{200}=\frac{995}{200}=\frac{199}{40}\)

\(B=\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{199\times200}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{199}-\dfrac{1}{200}\)

\(=1-\dfrac{1}{200}=\dfrac{199}{200}\)

E=0,5 x 199/200=199/400

\(E=\dfrac{0.5}{1.2}+\dfrac{0.5}{2\cdot3}+...+\dfrac{0.5}{199\cdot200}\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{200}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{199}{200}=\dfrac{199}{400}\)

\(A=\dfrac{11}{1.2}+\dfrac{11}{2.3}+\dfrac{11}{3.4}+...+\dfrac{11}{199.200}\)

\(A=11\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{199.200}\right)\)

\(A=11\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{199}-\dfrac{1}{200}\right)\)

\(A=11\left(1-\dfrac{1}{200}\right)\)

\(A=11.\dfrac{199}{200}=\dfrac{2189}{200}\)

\(B=3-\dfrac{1}{10}-\dfrac{1}{40}-\dfrac{1}{88}-\dfrac{1}{154}\)

\(B=3-\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}\right)\)

\(B=3-\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}\right)\)

\(B=3-\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\right)\)

\(B=3-\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{14}\right)\)

\(B=3-\dfrac{3}{7}=\dfrac{18}{7}\)

\(\frac{1}{1.2}+\frac{1}{3.4}+..........+\frac{1}{199.200}\)

Đặt S là biểu thức trên

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

\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+........+\frac{1}{199}+\frac{1}{200}-\frac{2.1}{2}-\frac{2.1}{4}-..........-\frac{2.1}{198}\)

\(\Rightarrow S=1+........+\frac{1}{200}-1-\frac{1}{2}-.........-\frac{1}{99}\)

Từ đây làm tiếp nhé

\(\frac{1}{101}+\frac{1}{102}+.......+\frac{1}{200}\)

Bạn tham khảo ở đây: https://toantieuhocpl.violet.vn/entry/show/entry_id/10846323

Lời giải:
Gọi tổng trên là $A$

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

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

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

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