Tính tổng sau: 1/2+1/2^2+1/2^3+1/2^4+...+1/2^2016
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NN
5
HB
1 tháng 3 2017
=\(\frac{1}{1}\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+.......+\(\frac{1}{2016}\)-\(\frac{1}{2017}\)+1
=\(\frac{1}{1}\)-\(\frac{1}{2017}\)+1
=\(\frac{2016}{2017}\)+1
=\(\frac{1}{2017}\)
1 tháng 3 2017
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2016.2017}+1\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2016}-\frac{1}{2017}+1\)
\(=1-\frac{1}{2017}+1\)
\(=\frac{2016}{2017}+1\)
\(=\frac{4033}{2017}\)
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
19 tháng 9 2015
A = 1+2+3+...+100
A = 100.(100+1):2 = 5050
1+2+3+.......+n = n(n+1):2
C = 2+4+6+.........+2016
C = (2 + 2016) x 1008 : 2 = 1017072
VP
1
KT
11 tháng 2 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(\Rightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(\Rightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Rightarrow\)\(A=1-\frac{1}{2^{2016}}\)
G
3
7 tháng 5 2017
\(S=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+4+...+2017}\)
\(S=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2035153}\)
\(S=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+....+\frac{2}{4070306}\)
\(S=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+....+\frac{2}{2017.2018}\)
\(S=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2017.2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(S=2.\left(\frac{1}{2}-\frac{1}{2018}\right)=2.\frac{504}{1009}=\frac{1008}{1009}\)
Vậy \(S=\frac{1008}{1009}\)
14 tháng 7 2015
a
so so hang
(100-1):1+1=100(so hang)
tong bang
(100+1)x100:2=5050
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\)
Ta có : \(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2017}}\right)\)
\(2A=2+\frac{2}{2}+\frac{2}{2^2}+\frac{2}{2^3}+...+\frac{2}{2^{2017}}\)
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\right)\)
\(A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{2016}}-\frac{1}{2^{2017}}\)
\(A=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)
Vậy \(A=\frac{2^{2018}-1}{2^{2017}}\)
4 tháng 5 2017
A=đã cho.
2A=1+1/2+1/2^2+1/2^3+...+1/2^2016.
2A-A=1-1/2^2017(khử).
A=1-1/2^2017.
KT
14 tháng 2 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2015}}+\frac{1}{2^{2016}}\)
\(\Leftrightarrow\)\(2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)
\(\Leftrightarrow\)\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
\(\Leftrightarrow\)\(A=1-\frac{1}{2^{2016}}\)
14 tháng 2 2018
\(A=\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{2016}}\)
\(2A=1+\frac{1}{2}+.........+\frac{1}{2^{2015}}\)
\(2A-A=\left(1+\frac{1}{2}+.....+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2016}}\right)\)
\(A=1-\frac{1}{2^{2016}}\)
Đặt A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\)
2A = \(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)
2A - A = \(\left(2+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)
A = \(2-\frac{1}{2^{2016}}\)