NT
2
K
Khách
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PT
3
15 tháng 2 2017
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)16
2A=\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2017}\)
2A-A=\(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+..+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)-\(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}+\frac{1}{2^{2017}}\)
A=\(\frac{1}{2017}-\frac{1}{2}\)
DT
0
VP
1
KT
11 tháng 2 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{2^{2016}}\)
PL
1
9 tháng 5 2019
\(\frac{B}{A}=\frac{\frac{2016}{1}+\frac{2015}{2}+...+\frac{2}{2015}+\frac{1}{2016}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\left(\frac{2016}{1}+1\right)+\left(\frac{2015}{2}+1\right)+...+\left(\frac{1}{2016}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{\frac{2017}{1}+\frac{2017}{2}+...+\frac{2017}{2016}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}\)
\(\frac{B}{A}=\frac{2017\cdot\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}+\frac{1}{2017}}=2017\div\frac{1}{2017}=4068289\)
NQ
0
Có : 2A = 1 + 1/2 + 1/2^2 +.....+ 1/2^2015
A = 2A - A = (1+1/2+1/2^2+.....+1/2^2015)-(1/2+1/2^2+.....+1/2^2016)
= 1 - 1/2^2016
Tk mk nha
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\right)\)
\(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2015}}-\frac{1}{2^{2016}}\)
\(A=1-\frac{1}{2^{2016}}\)