Tính nhanh:
M=\(\dfrac{1+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
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BP
30 tháng 8 2023
A= 1+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= \(\dfrac{2015}{2015}\)+(\(\dfrac{1}{2014}\)+1)+(\(\dfrac{2}{2013}\)+1)+...+(\(\dfrac{2013}{2}\)+1)
= 2015.(\(\dfrac{1}{2015}\)+\(\dfrac{1}{2014}\)+\(\dfrac{1}{2013}\)+...+\(\dfrac{1}{2}\))=2015.B
\(\Rightarrow\) \(\dfrac{A}{B}\)=2015
WT
1
21 tháng 4 2019
Đặt A = 1 + 2 + 22 + 23+ ...+ 22012
2A = 2 + 22 + 23 + 24 +....+22013
Lấy 2A - A = 2 + 22 +23 + 24 +....+22013 - 1-2-22- 23 - ... - 22012
A = 22013 - 1
Khi đó : M = A / 22014 -2
= 22013 - 1 / 2.( 22013 - 1 )
= 1/2
Vậy M= 1/2
DL
3
29 tháng 3 2016
Đặt A=1+2+22+...........+22012
2A=2+22+23+...........+22013
2A-A=(2+22+23+...........+22013)-(1+2+22+............+22012)
2A-A=22013-1
=>A=22013-1
Trở lại bài toán,ta có:
M=\(\frac{1+2+2^2+........+2^{2012}}{2^{2014}-2}\)
=\(\frac{2^{2013}-1}{2.2^{2013}-2}=\frac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\frac{1}{2}\)
Vậy M=\(\frac{1}{2}\)
A
6 tháng 8 2017
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{2}{2012}+\dfrac{1}{2013}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\left(1+\dfrac{2012}{2}\right)+\left(1+\dfrac{2011}{3}\right)+...+\left(1+\dfrac{2}{2012}\right)+\left(1+\dfrac{1}{2013}\right)+1\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2012}+\dfrac{2014}{2013}+\dfrac{2014}{2014}\)
\(\Leftrightarrow\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}+\dfrac{1}{2014}\right)\)
\(\Leftrightarrow x=\dfrac{2014.\left(\dfrac{1}{2}+\dfrac{1}{3}+....+\dfrac{1}{2014}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}}\)
\(\Leftrightarrow x=2014\)
Vậy \(x=2014\)
MV
6 tháng 8 2017
\(VP=\dfrac{2013}{1}+\dfrac{2012}{2}+...+\dfrac{1}{2013}\\ =\dfrac{2012}{2}+1+\dfrac{2011}{3}+1+...+\dfrac{1}{2013}+1+1\\ =\dfrac{2014}{2}+\dfrac{2014}{3}+...+\dfrac{2014}{2013}+\dfrac{2014}{2014}\\ =2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)x=2014\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2014}\right)\\ x=2014\)
Vậy x = 2014
LC
1
K
25 tháng 3 2018
Đặt A = 1 + 2 + 22 + 23 + ... + 22012
2A = 2 (1 + 2 + 22 + 23 + ... + 22012)
2A = 2 + 22 + 23 + 24 + ... + 22013
2A - A = (2 + 22 + 23 + 24 + ... + 22013) - (1 + 2 + 22 + 23 + ... + 22012)
=> A = 22013 - 1
Quay lại bài toán, ta có :
\(M=\dfrac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}=\dfrac{2^{2013}-1}{2^{2014}-2}=\dfrac{2^{2013}-1}{2\left(2^{2013}-1\right)}=\dfrac{1}{2}\)
DC
3
BA
9 tháng 4 2016
\(M=\frac{1+2+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
\(=>2M=\frac{2+2^2+2^3+2^4+...+2^{2013}}{2^{2014}-2}\)
\(=>M=\frac{\left(2+2^2+2^3+2^4+...+2^{2013}\right)-\left(1+2+2^2+2^3+...+2^{2012}\right)}{2^{2014}-2}\)
\(=>M=\frac{2^{2013}-1}{2^{2014}-2}\)
BA
9 tháng 4 2016
https://coccoc.com/search/math#query=%5B(2%5E2013)-1%5D%2F%5B(2%5E2014)-2%5D%5D
NT
1
DH
2
28 tháng 2 2015
Ta có:
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\)
Ta thấy :
\(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2}\)
\(\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(...\)
\(\frac{1}{2013.2014}=\frac{1}{2013}-\frac{1}{2014}\)
Ta có:
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1-\frac{1}{2014}\)
\(=\frac{2013}{2014}\)
\(M=\dfrac{1+2^2+2^3+...+2^{2012}}{2^{2014}-2}\)
\(=\dfrac{\left(1+2+2^2+2^3+...+2^{2012}\right)-2}{2^{2014}-2}\)
Gọi (1 + 2 + 22 + ... + 22012) - 2 là D
D = (1 + 2 + 22 + ... + 22012) - 2
2D = (2 + 22 + 23 + ... + 22013) - 4
D = (2 + 22 + 23 + ... + 22013) - 4 - (1 + 2 + 22 + ... + 22012) - 2
D = 22013 - 4 - 1 + 2
D = 22013 - 3
M = \(\dfrac{2^{2013}-3}{2^{2014}-2}\)
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