chứng minh rằng \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....+\frac{1}{2048}< 1\) 1
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T
4
11 tháng 8 2016
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{2048}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-...-\frac{1}{2048}+\frac{1}{2048}\)
\(=1-\frac{1}{2048}\)
\(=\frac{2047}{2048}\)
k mk nha
PN
4
23 tháng 4 2017
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
D
8 tháng 4 2017
Ta có: \(\frac{1}{2}-\frac{1}{4}=\frac{2}{4}-\frac{1}{4}=\frac{2-1}{4}=\frac{1}{4}\)
\(\frac{1}{8}-\frac{1}{16}=\frac{2}{16}-\frac{1}{16}=\frac{2-1}{16}=\frac{1}{16}\)
\(\frac{1}{32}-\frac{1}{64}=\frac{2}{64}-\frac{1}{64}=\frac{2-1}{64}=\frac{1}{64}\)
=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
=\(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}\)
=\(\frac{16}{64}+\frac{4}{64}+\frac{1}{64}=\frac{21}{64}\)
Ta có: \(\frac{21}{64}< \frac{21}{63}=\frac{1}{3}\)
=> \(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
8 tháng 4 2017
Đặt \(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}\)
\(2A=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\frac{1}{16}-\frac{1}{32}\)
\(A+2A=1-\frac{1}{64}\)
\(3A=1-\frac{1}{64}< 1\)
=>A<1/3
=>đpcm
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}+\frac{1}{2048}\)
=> 2A = \(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\)
=> 2A - A = \(\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{1024}+\frac{1}{2048}\right)\)
=> A = \(1-\frac{1}{2048}< 1\left(\text{ĐPCM}\right)\)