rút gọn biểu thức:1/2+1/2^2+1/2^3+....+1/2^100
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NL
1
HP
10 tháng 5 2016
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{100}}\)
=>\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
=>\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
=>\(A=1-\frac{1}{2^{100}}\)
PT
0
3 tháng 2 2019
a,M=2^0-2^1+2^2-2^3+2^4-2^5+.....+2^2012
2M=2^1-2^2+2^3-2^4+2^5-2^5+......-2^2012+2^2013
3M=2^0+2^2013
M=(2^0+2^2013)÷3
Vậy.......
b,N=3-3^2+3^3-3^4+3^5-3^6+.....+3^2011-3^2012
3N=3^2-3^3+3^4-3^5+3^6-3^7+......+3^2012-3^2013
4N=3-3^2013
N=(3-3^2013)÷4
Vậy........
K tao nhé ko lên lớp tao đánh m😈😈😈
NH
0
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
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
16 tháng 7 2016
Ta xét biểu thức sau :
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}\left[\left(\sqrt{n+1}\right)^2-\left(\sqrt{n}\right)^2\right]}\)
\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)(với n > 0)
Áp dụng : \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{100\sqrt{99}+99\sqrt{100}}\)
\(=\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}\right)+\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}\right)+...+\left(\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}\right)\)
\(=1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{99}}-\frac{1}{\sqrt{100}}=1-\frac{1}{\sqrt{100}}=1-\frac{1}{10}=\frac{9}{10}\)
PL
5 tháng 9 2018
Ta có : \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}\)
=> 2A - A = 1 - \(\frac{1}{2^{100}}\)
<=> A = 1 - \(\frac{1}{2^{100}}\)
5 tháng 9 2018
\(A=\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}.\)
\(\Rightarrow2A=1+\frac{1}{2^1}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{100}}\)
\(A=1-\frac{1}{2^{100}}\)
TC
0
Đặt \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(\Rightarrow2A=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\right)=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(\Rightarrow A=2A-A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}=1-\dfrac{1}{2^{100}}\)