CMR: 2/3+8/9+26/27+...+3^n-1/3^n < n-1/2
Những câu hỏi liên quan
Cho A = \(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+......+\frac{3^n-1}{3^n}\) CMR A > n-\(\frac{1}{2}\)
\(A=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
\(=\frac{3-1}{3}+\frac{9-1}{9}+\frac{27-1}{27}+...+\frac{3^n-1}{3^n}\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)+\left(\frac{9}{9}-\frac{1}{9}\right)+\left(\frac{27}{27}-\frac{1}{27}\right)+.....+\left(\frac{3^n}{3^n}-\frac{1}{3^n}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+....+\frac{1}{3^n}\right)\)
\(=n-\left(\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{3^n}\right)\)
Bây giờ ta chỉ cần chứng minh:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^n}< \frac{1}{2}\) là xong!
Thật vậy:\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
\(3B=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^{n-1}}\)
\(\Rightarrow2B=1-\frac{1}{3^n}\)
\(\Rightarrow B=\frac{1}{2}-\frac{\frac{1}{3^n}}{2}< \frac{1}{2}\)
Ta có:\(A=n-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^n}\right)\)
\(>n-\frac{1}{2}\left(đpcm\right)\)(bất đẳng thức đổi chiều)
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cho A = 2/3 + 8/9 + 26/27 +...+ 3^n - 1 / 3^n. chứng minh a > n - 1/2
Cho A=2/3+8/9+26/27+...+3^n -1/3^n. Chứng minh A>n-1/2
cho A= 2/3+8/9+26/17+...+3^n-1/3^n
n-A= 1/3+1/3^2+1/3^3+...+1/3^n
3(n-A)= 3+1/3+1/3^2+..+1/3^n-1
3(n-A)-(n-A)=1-1/3^n
2(n-A)=1-1/3^n
n-A=1/2-2/3^n<1/2
=>n-A<1/2
=> A
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cho A=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
=> n-A=\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^n}\)
=>\(3\left(n-A\right)\)=\(1\)\(+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{3n-1}}\)
=> \(3\left(n-A\right)-\left(n-A\right)=2\left(n-A\right)=1-\frac{1}{3^n}\)
=>\(2\left(n-A\right)< 1\)
=>\(n-A< \frac{1}{2}\)
=> \(A< n-\frac{1}{2}\)
Deu la tui het do
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Sao lại là n-A thế bạn? n đã tìm đc đâu
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Xem thêm câu trả lời
Cho C = 2/3 + 8/9 + 26/27 + ... + 3^n-1/3^n
Chứng ming rằng : C > n - 1/2
\(C=\frac{3-1}{3}+\frac{3^2-1}{3^2}+...+\frac{3^n-1}{3^n}\)
\(=1-\frac{1}{3}+1-\frac{1}{3^2}+...+1-\frac{1}{3^n}\)
\(=1+1+...+1-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}\right)\)
\(=n-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}\right)=n-D\)
\(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^n}\)
\(3D=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{n-1}}\)
\(\Rightarrow2D=1-\frac{1}{3^n}\Rightarrow D=\frac{1}{2}-\frac{1}{2.3^n}\)
\(\Rightarrow C=n-\left(\frac{1}{2}-\frac{1}{2.3^n}\right)=n-\frac{1}{2}+\frac{1}{2.3^n}>n-\frac{1}{2}\)
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Cho A = 2/3 + 8/9 + 26/27 + ...+ 3^n - 1/ 3^n. Chứng minh A> n - 1/2
cho A=2/3+8/9+26/27+......+3^n -1/3^n. chứng minh A>n-1/2
dễ mà các bn
1 CMR:
B=\(\frac{4}{3}+\frac{7}{3^2}+\frac{10}{3^3}+.....+\frac{3n+1}{3^n}< \frac{11}{4}\)(n thuộc N*;n>3)
A=\(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+...+\frac{100}{3^{100}}< \frac{3}{4}\)
C=\(\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^{20}-1}{3^{20}}>19\frac{1}{2}\)
Có : \(3A=1+\frac{2}{3}+\frac{3}{3^2}+...+\frac{100}{3^{99}}\)
\(3A-A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow2A< 1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
Có: \(6A< 3+1+\frac{1}{3}+...+\frac{1}{3^{98}}\)
\(6A-2A< 3-\frac{1}{3^{99}}< 3\)
\(\Rightarrow4A< 3\Rightarrow A< \frac{3}{4}\)(đpcm)
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\(choA=\frac{2}{3}+\frac{8}{9}+\frac{26}{27}+...+\frac{3^n-1}{3^n}\)
Chứng minh rằng \(A< n-\frac{1}{2}\)
A=2/3+8/9+26/27+...+3n+1/3n , chứng minh A>n-1/2
F=4/3+7/32+10/33+...+3n+1/3n,chứng minh E<3/4
so sánh L=(1-1/4).(1-1/9).(1-1/16)....(1-1/20) với 1/21