So sánh : \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}với1\)
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So sánh : \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}với1\)
So sánh : \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}với1\)
A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
2A = \(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{99}}\)
A = 2A - A = \(1-\frac{1}{2^{100}}
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So sánh:
\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}với1-\frac{1}{2^{2010}}\)
Với mọi số tự nhiên n\(\ge\)2. So sánh
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}với1\)
Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)
Ta có:
\(A=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\)\(< \)\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\left(1\right)\)
Mà \(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)\cdot n}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\)
\(=1-\frac{1}{n}< 1\left(2\right)\)(đúng. vì \(n\ge2\))
Từ (1) và (2) \(\Rightarrow A< B< 1\Rightarrow A< 1\)
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Bài 1: So sánh:
\(\frac{2}{51}+\frac{2}{52}+\frac{2}{53}+.................+\frac{2}{98}+\frac{2}{99}+\frac{2}{100}với1\)
Bài 2: Tìm n thuộc N để mỗi biểu thức sau là STN:
a, \(A=\frac{4}{n-1}+\frac{6}{n-1}-\frac{3}{n-1}\)
b, \(B=\frac{2n+9}{n+2}-\frac{3n}{n+2}+\frac{5n+17}{n+2}\)
1. so sánh
A=\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+với1\)
B=\(1-\left(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)với\frac{1}{2}\)
C=\(1-\left(\frac{1}{5}+\frac{1}{11}+\frac{1}{10}+\frac{1}{9}+\frac{1}{59}+\frac{1}{58}+\frac{1}{57}\right)với\frac{1}{2}\)
Cho A = 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026}\) ; B = \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\) . So sánh \(\frac{A}{B}với1\frac{2013}{2014}\)
Cho A = 1+\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{4026}\) ; B = \(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{4025}\) . So sánh \(\frac{A}{B}với1\frac{2013}{2014}\)
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\). So sánh A với 2
Ta có
\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
\(2A=1+\frac{2}{2}+\frac{3}{2^2}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)
Suy ra \(A=2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
Đặt \(n=\frac{1}{2}\) thì \(A=1+n+n^2+...+n^{99}-\frac{100}{2^{100}}\)
Xét \(B=1+n+n^2+...+n^{99}\Leftrightarrow B.n=n+n^2+n^3+...+n^{100}\)
\(\Leftrightarrow B.n=\left(1+n+n^2+...+n^{99}\right)+\left(n^{100}-1\right)\)
\(\Leftrightarrow B.n=B+n^{100}-1\Leftrightarrow B\left(n-1\right)=n^{100}-1\Leftrightarrow B=\frac{n^{100}-1}{n-1}\)
Suy ra \(A=\frac{\frac{1}{2^{100}}-1}{\frac{1}{2}-1}-\frac{100}{2^{100}}=2\left(1-\frac{1}{2^{100}}\right)-\frac{100}{2^{100}}=-\frac{102}{2^{100}}+2< 2\)
Vậy A < 2
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