\(B=\left(1+\dfrac{1}{100}\right)\times\left(1+\dfrac{1}{99}\right)\times....\times\left(1+\dfrac{1}{3}\right)\times\left(1+\dfrac{1}{2}\right)\)

\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)

\(B=\dfrac{101\times100\times....\times4\times3}{100\times99\times....\times3\times2}\)

\(B=\dfrac{101}{2}\)

\(\Rightarrow B=\left(\dfrac{100}{100}+\dfrac{1}{100}\right)\times\left(\dfrac{99}{99}+\dfrac{1}{99}\right)\times...\times\left(\dfrac{3}{3}+\dfrac{1}{3}\right)\times\left(\dfrac{2}{2}+\dfrac{1}{2}\right)\)

\(B=\dfrac{101}{100}\times\dfrac{100}{99}\times...\times\dfrac{4}{3}\times\dfrac{3}{2}\)

\(B=\dfrac{101}{2}\)( triệt tiêu các mẫu, tử giống nhau)

\(M=1+\frac{1}{3}-\frac{1}{3^2}+\frac{1}{3^3}-\frac{1}{3^4}+......+\frac{1}{3^{19}}-\frac{1}{3^{20}}\)

\(\Rightarrow\frac{1}{3}M=\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+\frac{1}{3^4}-\frac{1}{3^5}+.......+\frac{1}{3^{20}}-\frac{1}{3^{21}}\)

\(\Rightarrow\frac{1}{3}M+M=1+\frac{1}{3}+\frac{1}{3}-\frac{1}{3^{21}}\)

\(\Rightarrow\frac{4}{3}M=\frac{5}{3}-\frac{1}{3^{21}}\)\(\Rightarrow M=\frac{\frac{5}{3}-\frac{1}{3^{31}}}{\frac{4}{3}}\)

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

a,  ( 1/2 + 1) . ( 1/3 + 1) . (1/4 + 1) ... ( 1/999 + 1)

= 3/2 . 4/3 . 5/4 . 1000/999

= 1/2 . 1/1 . 1/1 ... 1000/1

= 1000/2 

= 500

b, (1/2-1) . (1/3-1) . (1/4-1) ... (1/1000-1)

= -1/2 . (-2)/3 . (-3)/4 ... (-999)/1000

= (-1)/1 . (-1)/1 . (-1)/1 ... (-1)/1000

= (-1)/1000

c, 3/2^2 . 8/3^2 . 15/4^2 ... 99/10^2

= 1.3/2.2 * 2.4/3.3 * 3.5/4.4***9.11/10.10

=( 1.2.3...99).(3.4.5...11)/(2.3.4....10).(2.3.4...10)

= 1.11/2.10

= 11/20

Ta có : 

\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{2003}\right)\left(1-\frac{1}{2004}\right)\)

\(B=\frac{1}{2}.\frac{2}{3}.....\frac{2002}{2003}.\frac{2003}{2004}\)

\(B=\frac{1.2.....2002.2003}{2.3.....2003.2004}\)

\(B=\frac{1}{2004}\)

Vậy \(B=\frac{1}{2004}\)

Chúc bạn học tốt ~ 

B= 1/2.2/3.3/4......2002/2003.2003/2004=1/2004