Bx3=1x3/1x2x3x4+1x3/2x3x4x5+...+1x3/97x98x99x100

Bx3=3/1x2x3x4+3/2x3x4x5+...+3/97x98x99x100

Bx3=1/1x2x3-1/2x3x4+1/2x3x4-1/3x4x5+...+1/97x98x99-1/98x99x100

BX3=1/1x2x3-1/98x99x100

BX3=1/6-1/970200

Bx3=161700/970200-1/970200

Bx3=161399/970200

B=161699/970200:3

B=161699/970200x1/3

B=161699/2910600

\(B=\frac{1}{1.2.3.4}+\frac{1}{2.3.4.5}+....+\frac{1}{97.98.99.100}\)

\(B=\frac{4-1}{1.2.3.4}+\frac{5-2}{2.3.4.5}+...+\frac{100-97}{97.98.99.100}\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{2.3.4}+\frac{1}{2.3.4}-\frac{1}{3.4.5}+...+\frac{1}{97.98.99}-\frac{1}{98.99.100}\right)\)

\(B=\frac{1}{3}\cdot\left(\frac{1}{1.2.3}-\frac{1}{98.99.100}\right)\)

\(B=\frac{1}{3}\cdot\frac{161699}{970200}=\frac{161699}{2910600}\)

1. C = 100 - 99 + 98 - 97 + ... + 2 - 1

C = ( 100 - 99 ) + ... + ( 98 - 97 )

C có 50 cặp số

C = 1 . 50

C = 50

2. Số thứ 50 của dãy số đó là :

4 . 49 + 1 = 197 

1.

C=100-99+98-97+...+2-1

=>C=(100-99)+(98-97)+...+(2-1)

=> C có 50 cặp

=> C=1+1+...+1

=> C=1.50

C = 50

thế còn bài thứ 2

(1-1/97)x(1-1/98)x...x(1-1/1000)

=96/97x97/98x...x999/1000

=96/1000

=0,096

bai toan @gmail.com

(1-1/97) x (1-1/98) x (1-1/99) x (1-1/1000) = 96/97 x 97/98 x 98/99 x 999/1000 

= 96 x 97 x 98 x 999 / 97 x 98 x 99 x 1000 = 12 x 111 / 11 x 125 = 1332 / 1375

\(C=\left(1-\frac{1}{97}\right).\left(1-\frac{1}{98}\right).\left(1-\frac{1}{99}\right)......\left(1-\frac{1}{1000}\right)\)

\(C=\frac{96}{97}.\frac{97}{98}.\frac{98}{99}........\frac{999}{1000}\)

\(C=\frac{96}{1000}\)

\(C=\frac{12}{125}\)

ket quả bằng 12 phần 125

hmmmmmmmmmmmmmmmmmmmmmmm