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\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+....+\left(\frac{1}{2010}+1\right)}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+....+\frac{2011}{2010}}\)
\(=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\right)}\)
\(=\frac{1}{2011}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+......+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+....+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}}{\left(\frac{2009}{2}+1\right)+\left(\frac{2008}{3}+1\right)+...+\left(\frac{1}{2010}+1\right)+1}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}+\frac{2011}{2011}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}}{2011\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}\right)}\)
\(A=\frac{1}{2011}\)
Ta có:
\(a_2^2=a_1.a_3;a_3^2=a_2.a_4;...;a^2_{2010}=a_{2009}.a_{2011}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3};\frac{a_2}{a_3}=\frac{a_3}{a_4};...;\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=\frac{a_3}{a_4}=...=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=...=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}=\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}\) (1)
Ta lại có:
\(\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_1}{a_2}.\frac{a_1}{a_2}...\frac{a_1}{a_2}=\frac{a_1}{a_2}.\frac{a_2}{a_3}...\frac{a_{2009}}{a_{2010}}.\frac{a_{2010}}{a_{2011}}=\frac{a_1}{a_{2011}}\) (2)
Từ (1) và (2) ta suy ra
\(\frac{a_1^{2010}+a_2^{2010}+...+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+...+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\)
Ta có :
\(a_2^2=a_1.a_3\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}\)
\(a^2_3=a_2.a_4\Rightarrow\frac{a_2}{a_3}=\frac{a_3}{a_4}\)
\(............\)
\(a^2_{2010}=a_{2009}.a_{2011}\Rightarrow\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
\(\Rightarrow\frac{a_1}{a_2}=\frac{a_2}{a_3}=........=\frac{a_{2009}}{a_{2010}}=\frac{a_{2010}}{a_{2011}}\)
Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=.......=\frac{a_{2010}}{a_{2011}}=k\)
\(\Rightarrow a_1=a_2.k\)
\(\Rightarrow a_1=a_3.k^2\)
\(\Rightarrow a_1=a_4.k^3\)
\(...............\)
\(\Rightarrow a_1=a_{2011}.k^{2010}\)
\(\Rightarrow\frac{a_1}{a_{2011}}=k^{2010}\) (1)
Ta có : \(k^{2010}=\left(\frac{a_1}{a_2}\right)^{2010}=\left(\frac{a_2}{a_3}\right)^{2010}=...=\left(\frac{a_{2010}}{a_{2011}}\right)^{2010}=\frac{a_1^{2010}}{a_2^{2010}}=\frac{a_2^{2010}}{a_3^{2010}}=....=\frac{a_{2010}^{2010}}{a_{2011}^{2010}}\)
\(=\frac{a_1^{2010}+a_2^{2010}+a_3^{2010}+....+a^{2010}_{2010}}{a_2^{2010}+a_3^{2010}+a_4^{2010}+....+a_{2011}^{2010}}\) ( theo TC DTSBN ) (2)
Từ (1) ; (2) \(\Rightarrow\frac{a_1^{2010}+a_2^{2010}+....+a_{2010}^{2010}}{a_2^{2010}+a_3^{2010}+....+a_{2011}^{2010}}=\frac{a_1}{a_{2011}}\) (đpcm)
Bước1: Chứng minh: x>ln(1+x)>x-x^2/2 (khảo sát hàm lớp 12)
Bước2: Đặt A=1+1/2+1/3+...+1/N.
B=1+1/2^2+1/3^2+...+1/N^2.
C=1+1/1.2+1/2.3+...+1/(N-1).N
D=ln(1+1)+ln(1+1/2)+ln(1+1/3)+...
...+ln(1+1/N).
Bước 3: Nhận xét: 1/k(k+1)=1/k-1/(k+1)
suy ra C=2-1/N <2
Bước 4: Nhận xét ln(k+1)-lnk=ln(1+1/k)
suy ra D=ln(N+1)
Bước 5: Nhận xét B<C<2
Bước 6: Chứng minh A->+oo (Omerta_V đã CM)
Bước 7: Từ Bước1 suy ra:
A>D>A-1/2B>A-1.
Bước 8: Vậy A xấp sỉ D với sai số tuyệt đối bằng 1.
Mà A->+oo. Nên khi N rất lớn thì sai số tương đối có thể coi là 0.
Cụ thể hơn Khi N>2^k thì sai số tương đối < k/2
Vậy khi N lớn hơn 1000000 thì ta có thể coi A=ln(N+1).
vậy đáp án là 5