A=1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

Bằng 26527650 đúng 100000000000000000000000000000%

A=1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650 

26527650

A = 26527650

A = 1x2x3 + 2x3x4 +…+ 100x101x102
Nhân A với 4 ta có :
A x 4 = 1x2x3x4 + 2x3x4x 4 + 3x4x5x4 +…+100x101x102x4
A x 4 = 1x2x3x4 + 2x3x4x(5-1) + 3x4x5x(6-2) + ... + 100x101x102x(103 - 99)
A x 4 = 1x2x3x4 + 2x3x4x5 - 1x2x3x4 + 3x4x5x6 - 2x3x4x5 + ... + 100x101x102x103 - 99x100x1001x102
Sau khi cộng - trừ giản ước ta có : A x 4 = 100x101x102x103
A = 100 x101x102x103 : 4 = 26527650

Ta có:

\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{100.101.102}\)

\(\Rightarrow\frac{1}{2}A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{100.101.102}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\)

\(\Rightarrow\frac{1}{2}A=\frac{1}{1.2}-\frac{1}{101.102}=\frac{2575}{5151}\Leftrightarrow A=\frac{2575}{10302}\)

ta co:1/1*2*3=(1/1*2-1/2*3):2
1/2*3*4=(1/1*2-1/2*3):2
...
cu nhu the cho den:
1/98*99*100=(1/98*99-1/99*100):2
suy ra : 1/1*2*3+1/2*3*4+1/3*4*5+...+1/98*99*100
=(1/1*2-1/2*3):2+(1/2*3-1/3*4):2+...+(1/98*99-1/99*100):2
=(1/1*2-1/2*3+1/2*3-1/3*4+...+1/98*99-1/99*100):2
=(1/1*2-1/99*100):2
=(1/2-1/9900)
=(4950/9000-1/9000):2
=4949/9000:2
=4949/18000
học tốt

ta có:
4s=1.2.3.(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+.........+k(k+1)(k+2)((k+3)-(k-1))
4s=1.2.3.4-1.2.3.0+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+........+k(k+1)(k+2)(k+3)-(k-1)k(k+1)(k+2)
4s=k(k+1)(k+2)(k+3)
ta biết rằng tích 4 số tự nhiên liên tiếp khi cộng thêm 1 luôn là 1 số chính phương
=>4s+1 là 1 số chính phương

\(A=\frac{1X2X3+2X4X6+4X8X12+8X16X24}{2X3X4+4X6X8+8X12X16+8X24X32}\)

\(A=\frac{1+1+1+1}{4+4+4+2}\)

\(A=\frac{4}{14}\)

CHÚC BẠN HỌC GIỎI !

A = 1.2.3 + 2.3.4 + ... + 20.21.22

⇒ 4A = 1.2.3.4 + 2.3.4.4 + ... + 20.21.22.4

= 1.2.3.4 + 2.3.4.(5 - 1) + 3.4.5.(6 - 2) ... + 20.21.22.(23 - 19)

= 1.2.3.4 - 1.2.3.4 + 2.3.4.5 - 2.3.4.5 + 3.4.5.6 + ... - 19.20.21.22 + 20.21.22.23

= 20.21.22.23

= 212520

⇒ A = 212520 : 4 = 53130