1/3 +(1/3)^2 +(1/3)^3 +....+(1/3)^2000
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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
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B= 3/2.4/3. ....2001/2000
B = 3.4....2001/2.3....2000
B =2001/2
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Ta có:
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=2001\)
bn cộng trên tử rồi thì phải trừ đi chứ ko phân số sẽ thay đổi
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Đặt A= 1 + 3 +32+33+.....+32000.
3A= 3 +32+33+.....+32001
3A - A = 3 +32+33+.....+32001 - (1 + 3 +32+33+.....+32000)
2A = 32001 - 1
=> A = \(\frac{3^{2001}-1}{2}\)
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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
Đặt A=1/3+1/3^2+1/3^3+...+1/3^2000
=>3A=1+1/3+1/3^2+ ...+1/3^1999
=>3A-A=(1+1/3+1/3^2+...+1/3^1999)-(1/3+1/3^2+1/3^3+...+1/3^2000)
=>2A=1-1/3^2000
=>A=(1-1/3^2000)/2