Cho a = 2 + 2 mũ 2 + 2 mũ 3 + chấm chấm chấm + 2 mũ 99 + 2 mũ 100 + 2 mũ 101 hỏi Nếu a chia cho 7 thì dư bao nhiêu?
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Úi gời cơi cộng chấm chấm chấm :)))
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=2.3+2^3.3+...+2^{2009}.3\)
\(A=3\left(2+2^3+...+2^{2010}\right)⋮3\)
-> Đpcm
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{2008}\left(1+2+2^2\right)\)
\(A=2.7+2^4.7+...+2^{2008}.7\)
\(A=7\left(2+2^4+...+2^{2008}\right)⋮7\)
-> Đpcm
![](https://rs.olm.vn/images/avt/0.png?1311)
2B= 22+23+24+...+2100
=>B=2B-B=22+23+24+...+2100-(21+22+23+...+299)=2100-2<2101-1
\(B=2^1+2^3+2^5+...+2^{99}\)
\(2^2B=2^2\left(2+2^3+2^5+...+2^{99}\right)\)
\(4B=2^3+2^5+2^7+...+2^{101}\)
\(4B-B=\left(2^3+2^5+2^7+...+2^{101}\right)-\left(2^1+2^3+2^5+..+2^{99}\right)\)
\(3B=2^{101}-2\)
\(B=\frac{2^{101}-2}{3}\) < \(F=2^{101}-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2^1+2^2+...+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2+2^2+2^3+...+2^{2010}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
A=2\(^1\)+2\(^2\)+...+2\(^{2010}\)
=(2\(^1\)+2\(^2\))+(2\(^3\)+2\(^4\))+...+(2\(^{2009}\)+2\(^{2010}\))
=2(1+2)+2\(^3\)(1+2)+...+2\(^{2009}\)(1+2)
=3(2+2\(^3\)+...+2\(^{2009}\))⋮3
![](https://rs.olm.vn/images/avt/0.png?1311)
A = 3 + 32 + 33 + ... + 3100
Số số hạng của A = ( 100 - 1 ) : 1 + 1 = 100 ssh . Ta chia A thanh 25 nhóm , mỗi nhóm cs 4 ssh .
=> A = ( 3 + 32 + 33 + 34 ) + .... + ( 397 + 398 + 399 + 3100 )
A = 3. ( 1 + 3 + 32 + 33 ) + .... + 397.( 1 + 3 + 32 + 33 )
A = 3. 40 + ... + 397 . 40
A = 40. ( 3 + ... + 397 )
=> A \(⋮\) 40 ( đpcm )
A = 3 + 32 + 33 + ... + 3100
Số số hạng của A = ( 100 - 1 ) : 1 + 1 = 100 ssh . Ta chia A thanh 25 nhóm , mỗi nhóm cs 4 ssh .
=> A = ( 3 + 32 + 33 + 34 ) + .... + ( 397 + 398 + 399 + 3100 )
A = 3. ( 1 + 3 + 32 + 33 ) + .... + 397.( 1 + 3 + 32 + 33 )
A = 3. 40 + ... + 397 . 40
A = 40. ( 3 + ... + 397 )
=> A 40 ( đpcm )
HT
dư 0 nhé