Tính nhanh
1/3+1/3^2+1/3^3+...+1/3^2007+1/3^2008
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\(M:N=\frac{\frac{2008}{1}+\frac{2007}{2}+\frac{2006}{3}+...+\frac{2}{2007}+\frac{1}{2008}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}\)
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
\(\Rightarrow M:N=\frac{2009.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2008}+\frac{1}{2009}}=2009\)
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
giải dùm mk với. chiều ni mk thi học kì. bài cuối sợ là bài này lắm
ta có tử số bằng :{2008 +2007/2 +... 2+1/2008} = {2007/2 +1 +2006/3+1 +...+1/2008+1} = {2009/2 +2009/3 +...+2009/2008} =
2009x{1/2 +1/3 +1/4+...+1/2009} . Vậy A = 2009
Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Rightarrow3A=3\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}\)
\(2A=3A-A\)
\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}-\frac{1}{3}-\frac{1}{3^2}-\frac{1}{3^3}-...-\frac{1}{3^{2007}}-\frac{1}{3^{2008}}\)
\(=1-\frac{1}{3^{2008}}\)
\(2A=1-\frac{1}{3^{2008}}\Rightarrow A=\frac{1-\frac{1}{3^{2008}}}{2}\)
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\)
\(\Leftrightarrow3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\)
\(\Leftrightarrow3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{2007}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2007}}+\frac{1}{3^{2008}}\right)\)
\(\Leftrightarrow2A=1-\frac{1}{3^{2008}}\)
\(\Leftrightarrow2A=\frac{3^{2008}-1}{3^{2008}}\)
\(\Leftrightarrow A=\frac{3^{2008}-1}{3^{2008}}\div2\)
\(\Leftrightarrow A=\frac{3^{2008}-1}{2.3^{2008}}\)