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

Ta có:

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

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

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

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

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

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Đặt A= 1 + 3 +3²+3^3+.....+3²⁰⁰⁰.

3A= 3 +3²+3^3+.....+3²⁰⁰¹

3A - A = 3 +3²+3^3+.....+3²⁰⁰¹ - (1 + 3 +3²+3^3+.....+3²⁰⁰⁰)

2A = 3²⁰⁰¹ - 1

=> A = \(\frac{3^{2001}-1}{2}\)

Ta có :\(3A=3+3^2+3^3+......+3^{2000}+3^{2001}\)

\(\Rightarrow3A-A=\left(3+3^2+3^3+....+3^{2000}+3^{2001}\right)-\left(1+3+3^2+.....+3^{2000}\right)\)

\(\Leftrightarrow2A=\left(3-3\right)+\left(3^2-3^2\right)+\left(3^3-3^3\right)+.....+\left(3^{2000}-3^{2000}\right)+\left(3^{2001}-1\right)\)

\(\Leftrightarrow2A=0+0+0+..+0+3^{2001}-1=3^{2001}-1\)

Vậy n=2001

a=1+3+3^2+....+3^2000

3a=3(1+3+3^2+....+3^2000)

3a=3+3^2+3^3+....+3^2001

3a-a=(3+3^2+3^3+....+3^2001)-(1+3+3^2+....+3^2000)

2a=3^2001-1(1)

Mà 2a=3^n-1.Từ (1)=>n=2001

Vậy n =2001

rkbgkl

3a=3+3²+3³+.............+3²⁰⁰¹

3a-a=3²⁰⁰¹-1

2a=3²⁰⁰¹-1

=>n=2001