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\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
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Đặt \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(A=\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\)
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\)
\(2A+A=\left(1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+\frac{1}{2^4}-\frac{1}{2^5}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+\frac{1}{2^5}-\frac{1}{2^6}\right)\)
\(3A=1-\frac{1}{2^6}\)
\(3A=\frac{2^6-1}{2^6}\)
\(A=\frac{\frac{2^6-1}{2^6}}{3}< \frac{1}{3}\)
Vậy \(A< 3\)
Chúc bạn học tốt ~
Ta có :
\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}< \frac{1}{3}\)
\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}< \frac{1}{3}\)
\(\frac{16+4+1}{64}< \frac{1}{3}\)
\(\frac{21}{64}< \frac{1}{3}\)
=> 1/2 - 1/4 + 1/8 - 1/16 + 1/32 - 1/64 < 1/3
đặt A bằng dãy trên
quy đồng mẫu số vs mẫu chung là 64. Ta có A=21/64<21/63=1/3
Chứng minh\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}<\frac{1}{3}\)
Đặt vế trái là A ta có
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(3A=2A+A=1-\frac{1}{64}<1\Rightarrow A<\frac{1}{3}\)
\(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{28}\)
=> \(\frac{16}{64}+\frac{8}{64}+\frac{4}{64}+\frac{2}{64}+\frac{1}{64}+\frac{1}{28}\)
=> \(\frac{31}{64}+\frac{1}{28}=>\frac{217}{448}+\frac{16}{448}=\frac{233}{448}\)
đúng 100%