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3 tháng 8 2019
\(P=...\)
\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)
\(=\frac{1}{99}-1=\frac{-98}{99}\)
3 tháng 8 2019
\(M=...\)
\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)
\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)
\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)
\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)
G
10 tháng 8 2017
Giải:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}+\dfrac{1}{2^7}\)
Lấy vế trừ vế, ta được:
\(A-\dfrac{1}{2}A=\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)
\(\Leftrightarrow\dfrac{1}{2}A=\dfrac{1}{2}-\dfrac{1}{2^7}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}-\dfrac{1}{2^7}}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{1}{2}\left(1-\dfrac{1}{2^6}\right)}{\dfrac{1}{2}}\)
\(\Leftrightarrow A=1-\dfrac{1}{2^6}\)
Vậy \(A=1-\dfrac{1}{2^6}\).
Chúc bạn học tốt!!!
10 tháng 8 2017
Đặt:
\(A=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+\dfrac{1}{64}\)
\(A=\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\)
\(2A=2\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)
\(2A=1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\)
\(2A-A=\left(1+\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}\right)-\left(\dfrac{1}{2^1}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+\dfrac{1}{2^5}+\dfrac{1}{2^6}\right)\)
\(A=1-\dfrac{1}{2^6}=1-\dfrac{1}{64}=\dfrac{63}{64}\)
26 tháng 5 2018
\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}+\frac{1}{64}-\frac{1}{128}\)
\(=1-\frac{1}{128}\)
\(\frac{127}{128}\)
HN
8 tháng 5 2017
Sửa đề:
\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< 1\)
Ta có:
\(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}=\dfrac{1}{4}+\dfrac{1}{16}+\dfrac{1}{64}\)
\(< \dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{3}{4}< \dfrac{4}{4}< 1\)
1 tháng 5 2019
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}\)
\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\)
\(A=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+.....+\frac{1}{7^2}\right)\)
\(< \frac{1}{2^2}\left(1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2^2}\cdot\frac{6}{7}\)
\(=\frac{3}{14}\)
\(< \frac{1}{2}\)
29 tháng 4 2018
\(N=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)
\(N=\dfrac{1}{2^1}-\dfrac{1}{2^2}+\dfrac{1}{2^3}-\dfrac{1}{2^4}+\dfrac{1}{2^5}-\dfrac{1}{2^6}\)
\(2N=1-\dfrac{1}{2^1}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}-\dfrac{1}{2^5}\)
\(2N+N=1-\dfrac{1}{2^6}\)
\(N=\dfrac{1}{3}-\dfrac{1}{2^6.3}< \dfrac{1}{3}\left(đpcm\right)\)
4 tháng 4 2018
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 ~
\(F=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^6}\)
\(\Rightarrow2F=1+\frac{1}{2}+....+\frac{1}{2^5}\)
\(\Rightarrow2F-F=F=\left(1+\frac{1}{2}+....+\frac{1}{2^5}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^6}\right)\)
\(\Rightarrow F=1-\frac{1}{2^6}\)