A=1.2+2.3+...+49.50

=>3A=1.2.3+2.3.3+...+49.50.3

=>3A=1.2.(3-0)+2.3.(4-1)+....+49.50.(51-48)

=>3A=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+49.50.51-48.49.50

=>3A=49.50.51

=>A=49.25.51=62475

=>3A=

Đặt A=1.2+2.3+3.4+4.5+...+49.50

3A=1.2.3+2.3.3+3.4.3+4.5.3+...+49.50.3

3A=1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+49.50.(51-48)

3A=1.2.3-0.1.2+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+49.50.51-48.49.50

3A=(1.2.3+2.3.4+3.4.5+4.5.6+...+49.50.51)-(0.1.2+1.2.3+2.3.4+3.4.5+...+48.49.50)

3A=49.50.51-0.1.2

3A=49.50.51

A=49.50.17

A=41650

đặt S làm tên biểu thức trên

ta có :

S =  1 .2 + 2.3 + 3.4 +4.5+.............+49 . 50

3S = 1.2.3 + 2.3.3 + 3.4.3 + 4.5.4 + ... + 49.50.3

3S = 1.2.3 + 2.3.(4-1) + 3.4.(5-2) + 4.5.(6-3 ) + ... + 49.50.(51 - 48 )

3S = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + 4.5.6 - 3.4.5 + ... + 49.50.51 - 48.49.50

3S = 49.50.51

S = 49.50.51 : 3

S = 41650

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

S = 1 + 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰⁰

2S = 2 + 2² + 2³ + 2⁴ + ... + 2¹⁰¹

S = 2S - S

= (2 + 2² + 2³ + ... + 2¹⁰¹) - (1 + 2 + 2² + ... + 2¹⁰⁰)

= 2¹⁰¹ - 1

------------

S = 1.2 + 2.3 + 3.4 + ... + 99.100 + 100.101

3S = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98) + 100.101.(102 - 99)

= 1.2.3 - 1.2.3 + 2

3.4 - 2.3.4 + 3.4.5 - ... - 98.99.100 + 99.100.101 - 99.100.101 + 100.101.102

= 100.101.102

S = 100 . 101 . 102 : 3

= 343400

------------

Q = 1² + 2² + 3² + ... + 100² + 101²

= 101.102.(2.101 + 1) : 6

= 348551

ĐKXĐ: \(x\ne0;x\ne-1\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2008}{2010}\)

\(\Leftrightarrow2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+..+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)(Biết công thức này chứ?)

\(\Leftrightarrow2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2008}{2010}\)

\(\Leftrightarrow1-\dfrac{2}{x+1}=\dfrac{2008}{2010}\)

\(\Leftrightarrow\dfrac{x-1}{x+1}=\dfrac{2008}{2010}\Leftrightarrow2010x-2010=2008x+2008\Leftrightarrow x=2009\left(tm\right)\)

Vậy x = 2009

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(A=2.\frac{12}{25}=\frac{2.12}{25}=\frac{24}{25}\)

cảm ơn bạn . kb với mik nha