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Khách
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4 tháng 10 2021
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
MH
4 tháng 10 2021
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
PV
18 tháng 4 2016
b) B = 2100 - 299 + 298 - 297 + ...+ 22 - 2
=> B x 2 = 2101 - 2100 + 299 - 298 + ...23 - 22
=> B x 2 + B = (2101 - 2100 + 299 - 298 + ...23 - 22 ) + (2100 - 299 + 298 - 297 + ...+ 22 - 2)
<=> B x 3 = 2101 - 2 = 2. ( 299 - 1)
=> B = \(\frac{2.\left(2^{99}-1\right)}{3}\)
Phần c) Làm tương tự Lấy C x 3 rồi + với C.
NT
1
NT
1
23 tháng 9 2017
Ta có : \(C=1+2^2+2^4+...+2^{100}\)
\(C.2^2=2^2\left(1+2^2+2^4+...+2^{100}\right)\)
\(C.4=2^2+2^4+2^6+...+2^{102}\)
\(C.4-C=2^{102}-1\)
\(C.3=2^{102}-1\)
\(C=\frac{2^{102}-1}{3}\)
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: \(\frac{49}{58}\cdot\frac{2^5}{4^2}-\frac{7^2}{-58}\cdot3\)
\(=\frac{49}{58}\cdot2-\frac{49}{58}\cdot3\)
\(=-1\cdot\frac{49}{58}\)
\(=-\frac{49}{58}\)
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)