Ta có:

\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)

\(\Leftrightarrow\frac{A}{B}=2001\)

vào câu hỏi tương tự nhé

A = ( 1 - 1/2 ) . ( 1 - 1/3 ) . ( 1 - 1/4 ) . ... . ( 1 - 1/2000)

A =  ( 2/2 - 1/2 ) . ( 3/3 - 1/3 ) . ( 4/4 - 1/4 ) . ... . ( 2000/2000 - 1/2000 )

A = 1/2 . 2/3 . 3/4 . ... . 1999/2000

A = 1.(2.3. ... . 1999)/ (2.3.4. ... .1999).2000

A = 1/2000

B = ( 1 + 1/2 ).(1 + 1/3 ).( 1+ 1/4 ). ... .(1+1/2000)

B = ( 2/2 + 1/2 ).(3/3+1/3).(4/4+1/4). ... .(1+1/2000)

B = 3/2.4/3.5/4. ... .2001/2000

B = (3.4.5. ... .2000).2001/2.(3.4. ... .2000)

B = 2001/2

B = 1000,5

A=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2000}\right)\)

A=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{1999}{2000}\)

A=\(\frac{1.2.3.4...1999}{2.3.4.5...2000}\)

A=\(\frac{1}{2000}\)

B=\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{2000}\right)\)

B=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2001}{2000}\)

B=\(\frac{3.4.5...2001}{2.3.4...2000}\)

B=\(\frac{2001}{2}\)

=1/2 . 2/3 ....1999/2000

=1.2....1999/2.3...2000

1/2000

B= 3/2.4/3. ....2001/2000

B = 3.4....2001/2.3....2000

B =2001/2

ko

A = 1 - 2 + 3 - 4 + 5 - 6  +......+ 1999 - 2000

Xét dãy số : 1; 2; 3; 4; 5; 6;......: 1999; 2000

Dãy số này là dãy số cách đều khoảng cách là: 2 - 1 = 1

Số số hạng của dãy số là : ( 2000 - 1) : 1 + 1 = 2000

            2000 : 2 = 1000

Vậy ta nhóm 2 số hạng liên tiếp của tổng A thành một nhóm thì tổng A có số nhóm là 1000 

Khi đó :

A = ( 1 - 2) + ( 3 - 4 ) + ( 5-6 ) +...( 1999 - 2000)

A = -1 x 1000

A =  - 1000

=2666666000

Có công thức như sau

1x2+2x3+3x4+...+nx(n+1)=nx(n+1)x(n+2):3

phần sau thì sao

a)-25

b)1000

minh nghi vay

\(a,A=2^0+2^1+2^2+....+\)\(2^{2010}\)

\(\Rightarrow2A=2^1+2^2+2^3+....+2^{2011}\)

 \(2A-A=\left(2^1+2^2+2^3+...+2^{2011}\right)-\left(2^0+2^1+2^2+...+2^{2010}\right)\)

  \(A=2^{2011}-2^0\)

\(A=2^{2011}-1\)

\(b,B=1+3+3^2+...+3^{100}\)

\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)

\(3B-B=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)

\(2B=3^{101}-1\)

\(\Rightarrow B=\frac{3^{101}-1}{2}\)

\(c,C=4+4^2+4^3+...+4^n\)

\(\Rightarrow4C=4^2+4^3+4^4+...+4^{n+1}\)

\(4C-C=\left(4^2+4^3+4^4+...+4^{n+1}\right)-\left(4+4^2+4^3+...+4^n\right)\)

\(3C=4^{n+1}-4\)

\(\Rightarrow C=\frac{4^{n+1}-4}{3}\)

\(d,D=1+5+5^2+...+5^{2000}\)

\(\Rightarrow5D=5+5^2+5^3+...+5^{2001}\)

\(5D-D=\left(5+5^2+5^3+...+5^{2001}\right)-\left(1+5+5^2+...+5^{2000}\right)\)

\(4D=5^{2001}-1\)

\(\Rightarrow D=\frac{5^{2001}-1}{4}\)

b)

B=1+3+3^2+3^3+..+3^100

=> 3B = 3 + 3^2 + 3^3 + ...+ 3^101

=> 3B - B = ( 3 + 3^2 + 3^3 + ...+ 3^101) - (1+3+3^2+3^3+..+3^100)

=> 2B = 3^101 - 1

=> B =( 3^101 - 1) / 2