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Bài 1:
a: \(2A=2^{101}+2^{100}+...+2^2+2\)
\(\Leftrightarrow A=2^{100}-1\)
b: \(3B=3^{101}+3^{100}+...+3^2+3\)
\(\Leftrightarrow2B=3^{100}-1\)
hay \(B=\dfrac{3^{100}-1}{2}\)
c: \(4C=4^{101}+4^{100}+...+4^2+4\)
\(\Leftrightarrow3C=4^{101}-1\)
hay \(C=\dfrac{4^{101}-1}{3}\)
\(A=\frac{1}{2}-\frac{2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+...+\frac{99}{2^{99}}-\frac{100}{2^{100}}\)
\(\Rightarrow2A=1-\frac{2}{2}+\frac{3}{2^2}-\frac{4}{2^3}+\frac{5}{2^4}-\frac{6}{2^5}+\frac{7}{2^6}-...+\frac{99}{2^{98}}-\frac{100}{2^{99}}\)
Cộng vế theo vế ta được: \(3A=1+\left(\frac{1}{2}-\frac{2}{2}\right)+\left(-\frac{2}{2^2}+\frac{3}{2^2}\right)+\left(\frac{3}{2^3}-\frac{4}{2^3}\right)+\left(-\frac{4}{2^4}+\frac{5}{2^4}\right)+...+\left(\frac{99}{2^{99}}-\frac{100}{2^{99}}\right)-\frac{100}{2^{100}}\)
\(\Rightarrow3A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+...+\frac{1}{2^{98}}-\frac{1}{2^{99}}-\frac{100}{2^{100}}\)
Xét \(B=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}+\frac{1}{2^{98}}-\frac{1}{2^{99}}\)
\(\Rightarrow2B=2-1+\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{97}}-\frac{1}{2^{98}}\)
Cộng vế theo vế ta được: \(3B=2+\left(1-1\right)+\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+...+\left(\frac{1}{2^{98}}-\frac{1}{2^{98}}\right)-\frac{1}{2^{99}}\)
\(\Rightarrow3B=2-\frac{1}{2^{99}}< 2\Rightarrow B< \frac{2}{3}\)
Mà \(3A=B-\frac{100}{2^{100}}\Rightarrow3A< B< \frac{2}{3}\Rightarrow A< \frac{2}{9}\)
\(\frac{1}{2}.A=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+\frac{5}{2^6}+...+\frac{100}{2^{101}}\)
\(A-\frac{1}{2}.A=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+...+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)(Do 3/2^3=1/2^2+1/2^3)
Đặt \(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(2B-B=1-\frac{1}{2^{100}}\)
Do đó:\(A-\frac{1}{2}A=1-\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\Rightarrow A=\frac{1}{2}-\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
\(=\frac{2^{99}-2-1}{2^{100}}\)
\(=\frac{2^{99}-3}{2^{100}}\)
Í nhầm cho mình sửa chỗ 1/2^101 thành 100/2^101 nhé
Mấy chỗ khác bạn từ đó sửa lại nha
Sorry
2A = 2^2 + 2^3 + 2^4 +...+ 2^101
=> 2A -A = (2^2 + 2^3 + 2^4 +...+ 2^101) - (2+ 2^2 + 2^3 + 2^4 +...+ 2^100)
=> A = 2^1010 - 2
A = 2 + 22 + 23 + 24 + .... + 2100
2A = 22 + 23 + 24 + ... + 2101
2A - A = 2101 - 2
=> A = 2101 - 2