tính:
D= \(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...-\frac{1}{2^{58}}\)
K
Khách
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Những câu hỏi liên quan
10 tháng 3 2017
\(\frac{1}{2^3}\)D= \(\frac{1}{2^4}-\frac{1}{2^7}+\frac{1}{2^{10}}-\frac{1}{2^{13}}+...+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)
D+ \(\frac{1}{2^3}\)D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^7}-\frac{1}{2^{10}}+\frac{1}{2^{10}}+...-\frac{1}{2^{58}}+\frac{1}{2^{58}}-\frac{1}{2^{61}}\)
\(\frac{9}{8}\)D= \(\frac{1}{2}-\frac{1}{2^{61}}\)=> D= \(\frac{\frac{1}{2}-\frac{1}{2^{61}}}{\frac{9}{8}}\)
ND
Tính D, biết
D=\(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+........-\frac{1}{2^{58}}\)
1
19 tháng 2 2019
\(D=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-\frac{1}{2^{10}}+...+\frac{1}{2^{55}}-\frac{1}{2^{58}}\)
\(\Rightarrow2^3D=2^2-\frac{1}{2}+\frac{1}{2^4}-\frac{1}{2^7}+....+\frac{1}{2^{52}}-\frac{1}{2^{55}}\)
\(\Rightarrow8D+D=2^2-\frac{1}{2^{58}}\)
\(\Rightarrow D=\frac{2^2-\frac{1}{2^{58}}}{9}\)
S
27 tháng 7 2019
\(2A=1-\frac{1}{2}+\frac{1}{2^2}-\frac{1}{2^3}+.....-\frac{1}{2^{99}}\Rightarrow2A+A=3A=\left(1-\frac{1}{2}+\frac{1}{2^2}-....-\frac{1}{2^{99}}\right)+\left(\frac{1}{2}-\frac{1}{2^2}+......-\frac{1}{2^{100}}\right)=1-\frac{1}{2^{100}}=\frac{2^{100}-1}{2^{100}}\Rightarrow A=\frac{2^{100}-1}{3.2^{100}}\)
\(2,4B=2+\frac{1}{2}+\frac{1}{2^3}+.....+\frac{1}{2^{97}}\Rightarrow4B-B=3B=\left(2+\frac{1}{2}+....+\frac{1}{2^{97}}\right)-\left(\frac{1}{2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\right)=2-\frac{1}{2^{99}}=\frac{2^{100}-1}{2^{99}}\Rightarrow B=\frac{2^{100}-1}{3.2^{99}}\)
\(3,C=\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\Rightarrow8C=4-\frac{1}{2}+\frac{1}{2^4}-.....-\frac{1}{2^{55}}\Rightarrow8C+C=9C=\left(4-\frac{1}{2}+\frac{1}{2^4}-....-\frac{1}{2^{55}}\right)+\left(\frac{1}{2}-\frac{1}{2^4}+\frac{1}{2^7}-....-\frac{1}{2^{58}}\right)=4-\frac{1}{2^{58}}=\frac{2^{60}-1}{2^{58}}\Rightarrow C=\frac{2^{60}-1}{9.2^{58}}\)
H
6 tháng 4 2019
a) 20,7 + 1,47 : 7 - 0,23 . 5
= 20,7 + 0,21 – 1,15
= 20,91 – 1,15
= 19,76
Ở trên vietjack có đó bn =)
6 tháng 4 2019
a, 20,7 + 1,47 : 7 - 0,23 . 5
=\(\frac{207}{10}+\frac{147}{100}:7-\frac{23}{100}.5\)
= \(\frac{207}{10}+\frac{21}{100}-\frac{23}{20}\)
= \(\frac{2091}{100}+\frac{-23}{20}\)
= \(\frac{494}{25}\)
CH
21 tháng 2 2019
D=1/2-1/24 +1/27-1/210 +.......+1/258
2D =1 - 1/2-1/24 +1/27-1/210 +.......+1/257
2D -D =(1 - 1/2-1/24 +1/27-1/210 +.......+1/257)-(1/2-1/24 +1/27-1/210 +.......+1/258)
D=1-1/258
10 tháng 10 2020
1) Đặt \(D=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)
\(\Rightarrow3D=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)
\(\Rightarrow3D-D=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(\Leftrightarrow2D=1-\frac{1}{3^{100}}\)
\(\Leftrightarrow D=\frac{3^{100}-1}{2\cdot3^{100}}\)
Vậy \(D=\frac{3^{100}-1}{2\cdot3^{100}}\)
2) Ta có: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)
\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)
\(=-1\cdot\frac{49}{58}\)
\(=-\frac{49}{58}\)
