A 

phân tích : 

= 2 + 6 + 12 + 20 + 30 ... + 2450

quy luật : 2 số liền nhau hơn kém nhau là các số chẵn liên tiếp :
   6 - 2 = 4 ; 12 - 6 = 6 ; 20 - 12 = 8

và bây giờ dùng tính chất dãy số để tính 

nhé !

A×3=1.2.3+2.3.3+3.4.3+.......+49.50.3

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

A×3=1.2.3-1.2.0+2.3.4-2.3.1+........+49.50.51-49.50.48

Ta thấy ngoài số 49.50.51 thì các số còn lại đều bị giản ước như 1.2.3 với 2.3.1;....nên 

A×3=49.50.51

A×3=124950

A=124950:3

A=41650.

Vậy A=41650.

S = 1/100

đặt A=1x2 + 3x4 + 5x6 + 7x8 + 9x10 + 11x12

     A=1x(3-1)+3x(5-1)+...+9x(11-1)+11x(13-1)

       =1x3-1+3x5-3+...+9x11-9+11x13-11

       =(1x3+3x5+...9x11+11x13)-(1+3+5+...+11)

đặt C=1x3+3x5+...+9x11+11x13

    6C=6x(1x3+3x5+...+9x11+11x13)

        =1x3x6+3x5x(7-1)+...+9x11x(13-7)+11x13x(15-9)

        =1x3x6+3x5x7-1x3x5+...+9x11x13-7x9x11+11x13x15-9x11x13

loại các số giống nhau ta được:

       =11x13x15+(1x3x6-1x3x5)

       =11x13x15+1x3x(6-5)

       =11x13x15+1x3x1

      =11x13x15+3

Vậy C=2538:6=423

=>A=423+1+3+5+...+11

      =423+36

Vậy A=459

Quá dễ

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

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 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101

A = 99x100x101 : 3

A = 333300

hok tốt

S= 1x2+2x3+3x4+4x5+...+ 20x21

3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )

3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3

3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)

3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21

3xS = 20x21x22

 S = 20x21x22 /3

S= 1x2+2x3+3x4+4x5+...+ 20x21

3xS=3x( 1x2+2x3+3x4+4x5+...+ 20x21 )

3xS = 1x2x3+2x3x3+3x4x3+....+20x21x3

3xS = 1x2x3 + 2x3x(4-1) + 3x4x(5-2)+........+20x21x(22-19)

3xS= 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 +......+20x21x22 - 19x20x21

3xS = 20x21x22

S = 20x21x22 /3

k mk nha

=>3D =1.2.3 + 2.3.3 + 3.4.3 + ..... + 99.100 .3

=> 3D = 1.2.3 - 2.3. ( 4-1) + 3.4. (5-2) + ... + 98.99 (100 - 97 ) + 99 . 100 . ( 101-98)

=> 3D= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +... + 98.99.100 -97.98.99 +99.100.101-98.99.100

=> 3D= 99.100.101

=> 3D= 999 900 

D= 999 900 .3 = 333 300

=>3D =1.2.3 + 2.3.3 + 3.4.3 + ..... + 99.100 .3

=> 3D = 1.2.3 - 2.3. ( 4-1) + 3.4. (5-2) + ... + 98.99 (100 - 97 ) + 99 . 100 . ( 101-98)

=> 3D= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 +... + 98.99.100 -97.98.99 +99.100.101-98.99.100

=> 3D= 99.100.101

=> 3D= 999 900 

D= 999 900 .3 = 333 300

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)

\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=\frac{1}{1}-\frac{1}{6}\)

\(=\frac{5}{6}\)

\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)

\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)

\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)

\(=\)\(\frac{5}{6}\)

Hok tốt

Gọi biểu thức trên là A, ta có :

A = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

A x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

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 + ... + 99x100x101 - 98x99x100.

A x 3 = 99x100x101

A = 99x100x101 : 3

A = 333300

Gọi biểu thức trên là S, ta có :

S = 1x2 + 2x3 + 3x4 + 4x5 + ...+ 99x100

S x 3 = 1x2x3 + 2x3x3 + 3x4x3 + 4x5x3 + ... + 99x100x3

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

S x 3 = 1x2x3 + 2x3x4 - 1x2x3 + 3x4x5 - 2x3x4 + 4x5x6 - 3x4x5 + ... + 99x100x101 - 98x99x100.

S x 3 = 99x100x101

S = 99x100x101 : 3

S = 333300

\(A=\frac{1.98+2.97+3.96+...+98.1}{1.2+2.3+3.4+...+98.99}=\frac{1.\left(100-2\right)+2\left(100-3\right)+3\left(100-4\right)+...+98\left(100-99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{1.100-1.2+2.100-2.3+3.100-3.4+...+98.100-98.99}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{\left(1.100+2.100+3.100+...+98.100\right)-\left(1.2+2.3+3.4+...+98.99\right)}{1.2+2.3+3.4+...+98.99}\)

\(A=\frac{100\left(1+2+3+...+98\right)}{1.2+2.3+3.4+...+98.99}-1\)

Ta có: 1+2+3+...+98=98.99:2=4851

Đặt B=1.2+2.3+3.4+...+98.99  => 3B=1.2.3+2.3.3+3.4.3+...+98.99.3 = 1.2.3+2.3.(4-1)+3.4(5-2)+...+98.99(100-97)

=> 3B=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+98.99.100-97.98.99 = 98.99.100

=> B=33.98.100. Thay vào A được:

\(A=\frac{100.4851}{33.98.100}-1=\frac{3}{2}-1=\frac{1}{2}\)

A=1/2