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\(A=\frac{\left(1+2+3+...+100\right)\left(\frac{1}{4}+\frac{1}{6}-\frac{1}{2}\right)\left(63.1,2-21.3,6+1\right)}{1-2+3-4+....+99-100}\)
\(=\frac{\frac{100\left(100+1\right)}{2}\left(\frac{3+2-6}{12}\right)\left[63\left(1,2-1,2\right)+1\right]}{\left(1-2\right)+\left(3-4\right)+....+\left(99-100\right)}\)
\(=\frac{5050.\left(-\frac{1}{12}\right).1}{-1+\left(-1\right)+\left(-1\right)+...+\left(-1\right)}\)
\(=\frac{2525.\left(-\frac{1}{6}\right)}{-50}=\frac{101}{12}\)
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 \(A=2-\frac{102}{2^{100}}\)
ta có: \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{100^2}=1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)\)
Lại có: \(\frac{1}{2^2}>\frac{1}{2.3};\frac{1}{3^2}>\frac{1}{3.4};\frac{1}{4^2}>\frac{1}{4.5};...;\frac{1}{100^2}>\frac{1}{100.101}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{100.101}\)
\(=\frac{1}{2}-\frac{1}{101}\)
\(\Rightarrow1-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\right)>1-\left(\frac{1}{2}-\frac{1}{101}\right)=1-\frac{1}{2}+\frac{1}{101}\)
\(=\frac{1}{2}+\frac{1}{101}\)
mà \(\frac{1}{2}=\frac{50}{100}>\frac{1}{100}\Rightarrow\frac{1}{2}+\frac{1}{101}>\frac{1}{100}\)
=> đ p c m
\(2A=2+\frac{3}{2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)
\(3E-E=2E=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
=>E=... tự tính
nobita kun ơi............em vừa phải thôi nhé. Đã không giúp con spam nữa. điều nay ai chả biết