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Ta có:
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(2A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\)
\(2A-A=\left(2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}\right)-\left(1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\right)\)
\(A=2+\frac{3}{2^2}+\frac{4}{2^3}+\frac{5}{2^4}+...+\frac{100}{2^{99}}-1-\frac{3}{2^3}-\frac{4}{2^4}-...-\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{2^4}\right)+...+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(A=1+\frac{3}{4}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
Đặt \(B=\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}\)
\(\Rightarrow A=1+\frac{3}{4}+B-\frac{100}{2^{99}}\) (1)
Ta có:
\(B=\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}...+\frac{1}{2^{99}}\)
\(\Rightarrow2B=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...+\frac{1}{2^{98}}\)
\(2B-B=\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\right)-\left(\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\right)\)
\(B=\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{98}}-\frac{1}{2^3}-\frac{1}{2^4}-...-\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(B=\frac{1}{2^2}+\left(\frac{1}{2^3}-\frac{1}{2^3}\right)+\left(\frac{1}{2^4}-\frac{1}{2^4}\right)+...+\left(\frac{1}{2^{98}}-\frac{1}{2^{98}}\right)-\frac{1}{2^{99}}\)
\(B=\frac{1}{4}+0+0+...+0-\frac{1}{2^{99}}\)
\(B=\frac{1}{4}-\frac{1}{2^{99}}\)
Từ (1)
\(\Rightarrow A=1+\frac{3}{4}+\left(\frac{1}{4}-\frac{1}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(A=\frac{7}{4}+\frac{1}{4}-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
\(A=2-\frac{2}{2^{100}}-\frac{100}{2^{100}}\)
\(A=2-\frac{102}{2^{100}}\)
Vậy \(A=2-\frac{102}{2^{100}}\)
Bài 1:
Ta có:
\(\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}\)
\(=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+...+\frac{19}{81.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{81}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Mà \(\frac{99}{100}< 1\)
\(\Rightarrow\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+...+\frac{19}{9^2.10^2}< 1\left(đpcm\right)\)
\(2A=2+\frac{3}{2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(3E-E=2E=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=>E=... tự tính
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\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+\frac{5}{2^5}+...+\frac{100}{2^{100}}\)
\(\Rightarrow2A=2+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(\Rightarrow A=1+\frac{3}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow B=2-\frac{1}{2^{99}}\Rightarrow A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
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