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17 tháng 4 2017
Ta có :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..............+\dfrac{1}{2^{2015}}+\dfrac{1}{2^{2016}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2015}}\)
\(\Rightarrow2A-A=\left(1+\dfrac{1}{2}+..........+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{2016}}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{2016}}\)
\(\Rightarrow A=\dfrac{2^{2016}-1}{2^{2016}}\)
~ Học tốt ~
DT
2
9 tháng 10 2018
M=2018^2-2017^2+2016^2-2015^2+............+2^2-1^2
M=(2018+2017).(2018-2017)+(2016+2015).(2016-2015)+...........+(2+1).(2-1)
M=2018+2017+2016+2015+.................+2+1
M=2018.(2018+1)/2=2018.2019/2
M=1009.2019M=2037171
CK
1
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
4 tháng 5 2016
\(2.TS=2+2^2+2^3+2^4+...+2^{2017}\)
\(TS=2.TS-TS=2^{2017}-1\)
\(B=\frac{2^{2017}-1}{1-2^{2016}}=-\frac{1-2.2^{2016}}{1-2^{2016}}=-\frac{1-2^{2016}-2^{2016}}{1-2^{2016}}=-\left(1-\frac{2^{2016}}{1-2^{2016}}\right)\)
YN
4 tháng 4 2022
`Answer:`
\(T=\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{2016}{2^{2015}}+\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T=2+\frac{3}{2}+\frac{4}{2^2}+...+\frac{2016}{2^{2014}}+\frac{2017}{2^{2015}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{3}{2}-\frac{2}{2}\right)+\left(\frac{4}{2^2}-\frac{4}{2^2}\right)+...+\left(\frac{2017}{2^{2015}}-\frac{2016}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow2T-T=2+\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
Ta đặt \(V=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
\(\Rightarrow T=2+V-\frac{2017}{2^{2016}}\text{(*)}\)
\(\Leftrightarrow2V=1+\frac{1}{2}+...+\frac{1}{2^{2014}}\)
\(\Leftrightarrow2V-V=\left(1+\frac{1}{2}+...+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)\)
\(\Leftrightarrow2V-V=1-\frac{1}{2^{2015}}\text{(**)}\)
Từ (*)(**)\(\Rightarrow T=2+\left(1-\frac{1}{2^{2015}}\right)-\frac{2017}{2^{2016}}\)
\(\Leftrightarrow T=3-\frac{1}{2^{2015}}-\frac{2017}{2^{2016}}\)
`=>T<3`
NQ
0
HN
13
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=\(\frac{1}{2017}-\frac{1}{2}\)
M = 1-2+22-23+...+22016
2M = 2-22+23-24+...+22017
3M = 2M + M = 1 + 22017
=> M = \(\frac{2^{2017}+1}{3}\)