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a)\(\frac{32}{64}-\frac{16}{64}+\frac{8}{64}-\frac{4}{64}+\frac{2}{64}-\frac{1}{64}\le\frac{1}{3}\)
\(\Rightarrow\frac{32-16+8-4+2-1}{64}=\frac{23}{64}\)\
\(\Rightarrow\frac{23}{64}=0,359375;\frac{1}{3}=0,33333...\)
đề sao lạ vậy
\(3B=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(B=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow4B=3B+B=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
+ Đặt \(M=1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{98}}-\frac{1}{3^{99}}\)
\(3M=3-1+\frac{1}{3}-\frac{1}{3^2}+...+\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(\Rightarrow4M=3M+M=3-\frac{1}{3^{99}}\)
\(\Rightarrow M=\frac{3}{4}-\frac{1}{3^{99}\cdot4}\)
\(\Rightarrow4B=M-\frac{100}{3^{100}}=\frac{3}{4}-\frac{1}{3^{99}\cdot4}-\frac{100}{3^{100}}\)
\(\Rightarrow B=\frac{3}{16}-\frac{1}{3^{99}\cdot16}-\frac{100}{3^{100}\cdot4}\) \(\Rightarrow B< \frac{3}{16}\)
a) \(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(\Rightarrow3A=2A+A=1-\frac{1}{2^6}\)
\(\Rightarrow A=\frac{1}{3}-\frac{1}{2^6\cdot3}< \frac{1}{3}\) ( đpcm )
Nhầm đầu bài nhoa:
Phải là \(-\frac{100}{3^{100}}\)
Đặt A = \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+....\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-\frac{4}{3^3}+....\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=A+3A=\left(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+.....\frac{1}{3^{98}}-\frac{1}{3^{99}}\right)\)
Đặt B = \(1-\frac{1}{3}+\frac{1}{3^2}-\frac{1}{3^3}+....\frac{1}{3^{98}}-\frac{1}{3^{99}}\Rightarrow4A< B\left(1\right)\)
\(\Rightarrow3B=3-1+\frac{1}{3}-\frac{1}{3^2}+....\frac{1}{3^{97}}-\frac{1}{3^{98}}\)
\(4B=B+3B=3-\frac{1}{3^{99}}< 3\Rightarrow4B< 3\Rightarrow B< \frac{3}{4}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow4A< B< \frac{3}{4}\Rightarrow4A< \frac{3}{4}\Rightarrow A< \frac{3}{4}:4\Rightarrow A< \frac{3}{4}.\frac{1}{4}\Rightarrow A< \frac{3}{16}\)
=> đpcm.
