a: \(\Leftrightarrow\dfrac{1}{3}A=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{n+1}}\)

\(\Leftrightarrow-\dfrac{2}{3}A=\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\)

hay \(A=\left(\dfrac{1}{3^{n+1}}-\dfrac{1}{3}\right):\dfrac{-2}{3}=\dfrac{1-3^n}{3^{n+1}}\cdot\dfrac{3}{-2}=\dfrac{3^n-1}{3^n\cdot2}\)

b: \(\dfrac{1}{3}B=\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{2009}}\)

\(\Leftrightarrow B\cdot\dfrac{-2}{3}=\dfrac{1}{3^{2009}}-\dfrac{1}{3}=\dfrac{1-3^{2008}}{3^{2009}}\)

\(\Leftrightarrow B=\dfrac{3^{2008}-1}{3^{2009}}:\dfrac{2}{3}=\dfrac{3^{2008}-1}{2\cdot3^{2008}}\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)

\(2A=3A-A\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)

\(=1-\frac{1}{3^{2008}}\)

\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)

\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)

\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)

\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)

\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)

\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)

\(\frac{M}{N}=\frac{\frac{1}{2007}+\frac{2}{2006}+......+\frac{2006}{2}+\frac{2007}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.......+\frac{1}{2006}+\frac{1}{2007}}\)

\(\frac{M}{N}=\frac{\frac{1}{2007}+1+\frac{2}{2006}+1+.......+\frac{2007}{1}+1+\frac{2008}{2008}-2008}{\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+.....+\frac{1}{2}}\)

\(\frac{M}{N}=\frac{\frac{2008}{2007}+\frac{2008}{2006}+....+\frac{2008}{1}+\frac{2008}{2008}-2008}{\frac{1}{2008}+........+\frac{1}{2}}\)

đến đây là ra rùi ha 

ê tớ chẳng hiểu gì cả

cậu làm tắt à

please cậu giúp tớ cả bài đi mà

Đặt A=1+a+a^2+a^3+...+a^n

a*A=a+a^2+a^3+a^4+...+aⁿ⁺¹

a*A+1=1+a+a^2+a^3+...+a^n+aⁿ⁺¹=A+aⁿ⁺¹

a*A-A=aⁿ⁺¹-1

(a-1)A=aⁿ⁺¹-1

A=(aⁿ⁺¹-1)/(a-1)

\(M:N=\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)

Ta có tử số bằng: 2008+2007/2+2006/3+2005/4+…..+2/2007+1/2008 
(Phân tích 2008 thành 2008 con số 1 rồi đưa vào các nhóm) 
= (1 + 2007/2) + (1 + 2006/3) + (1 + 2005/4) +... + (1 + 2/2007) + ( 1 + 1/2008) + (1) 
= 2009/2 + 2009/3 + 2009//4 + ……. + 2009/2007 + 2009/2008 + 2009/2009 
= 2009 x (1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009) 
Mẫu số: 1/2 + 1/3 + 1/4 + ... + 1/2007 + 1/2008 + 1/2009 
\(\Rightarrow M:N=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)

Cái bài 2 nhân với 1 là 2/2 nên nhân cả tử cả mẫu với 2 ra 6=2*3

                                                                                              12=3*4

                                                                                               .........

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giúp mình với !!!!!!!!!!!!!!!!!!!!!!!!

Câu b, bài b1 chứng minh \(a=2^{2006}-1?\)

A=2008+2007/2+2006/3+2005/4+...+2/2007+1/2008

1/2+1/3+1/4+1/5+...+1/2007+1/2008

=(1+2007/2)+(1+2006/3)+(1+2005/4)+...+(1+2/2007)+(1+1/2008)

1/2+1/3+1/4+...+1/2008

=2009(1/2+1/3+1/4+...+1/2008)

1/2+1/3+1/4+..+1/2008

=2009