![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1.A=2^2+2^4+...+2^2010
=> 2^2 A= 2^4+2^6+..+2^2012
=> 2^2 A - A=( 2^4+2^6+..+2^2012 ) -(2^2+2^4+...+2^2010 )
=> 3A= 2^2012 -2^2
=> A= (2^2012-2^2)/3
B=3-3^2+3^3-...-3^2010
=>3B= 3^2 -3^3+3^4-...-3^2011
=> 3B + B = (3^2 -3^3+3^4-...-3^2011) +(3-3^2+3^3-...-3^2010)
=> 4B =3-3^2011
=> B= (3-3^2011)/4
2.
A=3+3^2+..+3^100
=> 3A =3^2+3^3+...+3^101
=> 3A- A = (3^2+3^3+...+3^101)-(3+3^2+..+3^100)
=> 2A=3^101 -3
=> 2A+3 =3^101 mà 2A+3 =3^n
=> n=101
![](https://rs.olm.vn/images/avt/0.png?1311)
A=1+2+2^2+2^3+2^4+2^5
2A=2+2^2+2^3+2^4+2^5+2^6
2A-A=(2+2^2+2^3+2^4+2^5+2^6)-(1+2+2^2+2^3+2^4+2^5)
A=2^6-1
A=64-1
A=63
B=10+12+14+....+2010
B=(2010+10).1001:2
B=2020.1001:2
B=2022020:2
B=1011010
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(A=2^0+2^1+2^2+...+2^{2010}\)
\(\Rightarrow2A=2+2^2+2^3+...+2^{2011}\)
\(\Rightarrow2A-A=\left(2+...+2^{2011}\right)-\left(2^0+2^1+...+2^{2010}\right)\)
\(\Rightarrow2A-A=2^{2011}-2^0\)
\(\Rightarrow A=2^{2011}-1\)
b ) \(B=1+3+3^2+...+3^{100}\)
\(\Rightarrow3B=3+3^2+3^3+...+3^{101}\)
\(\Rightarrow3B-B=\left(3+3^2...+3^{2011}\right)-\left(1+3+...+3^{2010}\right)\)
\(\Rightarrow2B=3^{2011}-1\)
\(\Rightarrow B=\frac{3^{2011}-1}{2}\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Cho \(A=2^0+2^1+...+2^{2009}+2^{2010}\)
-> \(2A=2^1+2^2+...+2^{2010}+2^{2011}\)
\(2A-A=2^1+2^2+...+2^{2011}-\left(2^0+2^1+...+2^{2010}\right)\)
\(A=2^{2011}-2^0\)
\(A=2^{2011}-1\)
\(A=2+2^2+2^3+...+2^{2010}\)
\(2A=2^2+2^3+2^4+...+2^{2011}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2011}\right)-\left(2+2^2+2^3+...+2^{2010}\right)\)
\(A=2^{1011}-2\)
Đặt A = 21 + 22 + ... + 22010
=> 2A = 22 + 23 + ... + 22011
=> 2A - A = ( 22 + 23 + ... + 22011 ) - ( 21 + 22 + ... + 22010 )
A = 22011 - 2
VẬy A = 22011 - 2