Xin hãy giúp mình

a,=3/2*4/3*....100/99

=3*4*5*....*100/2*3*...*99

=100/2=50

b, nhân lên băng:

1*2*3*...*99/2*3*...*100=1/100

99 x ( 23 + 57 + 20 ) + ( 1 + 2 + 3 +...+ 100 ) = ( 100 - 1 ) +1 = 100

= 99 x 100 + [(100 + 1 ) x 100 : 2 ] 

= 9900 + 5050

= 14950

=  99 X ( 23 + 57 + 20 )  + {( 101 + 101 + ... + 101 ) CÓ 100 SỐ HẠNG }

= ( 99 X 100 ) + (101 X 100 : 2 )

 = 9900 + 5050 

= 14950

Ta đặt:

S = 1 + (1 + 2) + (1 + 2 + 3) + ... + (1 + 2 + 3 + ... + 99)

S x 2 = 1 x 2 + 2 x 3 + 3 x 4 + .... + 99 x 100

\(D=\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...\frac{1}{99.100}\)

\(D=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}...\frac{1}{99}-\frac{1}{100}\)

\(D=\frac{1}{1}-\frac{1}{100}\)

\(D=\frac{99}{100}\)

Vậy tổng D bằng \(\frac{99}{100}\)

tổng quát: \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

áp dụng ta có: \(D=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)

Sửa đề: 

\(\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)

\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{50}{100}-\dfrac{1}{100}=\dfrac{49}{50}\)