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![](https://rs.olm.vn/images/avt/0.png?1311)
A=4+4^1+4^2+..........+4^2004
A.3=4^2007-4
\(A=\frac{\left(4^{2007}-4\right)}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Chắc đặt nhầm lớp rồi
Ta có :\(B=4^{2004}+4^{2003}+...+4^2+4+1\)
\(4B=\left(4^{2004}+4^{2003}+...+4^2+4+1\right).4\)
\(4B=4^{2005}+4^{2004}+...+4^3+4^2+4\)
\(4B-B=\left(4^{2005}+4^{2004}+...+4^3+4^2+4\right)\)\(-\left(4^{2004}+4^{2003}+...+4+1\right)\)
\(3B=\left(4^{2005}-1\right)\)\(\Rightarrow\frac{4^{2005}-1}{3}\)
\(\Rightarrow A=75.\frac{4^{2005}-1}{3}+25\)
\(\Rightarrow A=25.\left(4^{2005}-1\right)+25\)
\(\Rightarrow A=25.\left(4^{2005}-1+1\right)\)
\(\Rightarrow A=25.4.4^{2004}\)
\(\Rightarrow A=100.4^{2004}\)
Mà 100 chia hết 100 nên \(100.4^{2004}\) chia hết cho 100
B=4^0 + 4^1 +...+ 4^2004
4B=4^1+4^2+...+4^2005
3B=4^2004-4^0
B=(4^2004-4^0):3
Thay B vào ta có :
A=75.(4^2004-4^0):3+25
A=25.(4^2004-4^0)+25
A=25.4^2004
A=100.4^2003
Vậy A chia hết cho 100
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=75.\left(4^{2004}+4^{2003}+......+4^2+1\right)+25\)
Đặt :
\(B=4^{2004}+4^{2003}+.......+4^2+4+1\)
\(\Leftrightarrow4B=4^{2005}+4^{2004}+........+4^2+4\)
\(\Leftrightarrow4B-B=\left(4^{2005}+4^{2004}+......+4^2+4\right)-\left(4^{2004}+4^{2003}+.....+4+1\right)\)
\(\Leftrightarrow3B=4^{2005}-1\)
\(\Leftrightarrow B=\dfrac{4^{2005}-1}{3}\)
\(\Leftrightarrow A=75.\dfrac{4^{2005}-1}{3}+25\)
\(\Leftrightarrow A=25.\left(4^{2004}-1+1\right)\)
\(\Leftrightarrow A=25.4.4^{2003}\)
\(\Leftrightarrow A=100.4^{2003}⋮100\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt B = 42004 + 42003 + ... + 42 + 4 + 1 (có 2005 số; 2005 chia 2 dư 1)
B = (42004 + 42003) + (42002 + 42001) + ... + (42 + 4) + 1
B = 42003.(4 + 1) + 42002.(4 + 1) + ... + 4.(4 + 1) + 1
B = 42003.5 + 42002.5 + ... + 4.5 + 1
B = 5.(42003 + 42002 + ... + 4) + 1 chia 5 dư 1
=> B = 5.k + 1 (k là số chia hết cho 4)
=> A = 75.(5.k + 1) + 25
=> A = 75.5k + 75 + 25
=> A = (...00) + 100
=> A = (...00) chia hết cho 100 (đpcm)
Đặt B = 42004 + 42003 + ... + 42 + 4 + 1 (có 2005 số; 2005 chia 2 dư 1)
B = (42004 + 42003) + (42002 + 42001) + ... + (42 + 4) + 1
B = 42003.(4 + 1) + 42002.(4 + 1) + ... + 4.(4 + 1) + 1
B = 42003.5 + 42002.5 + ... + 4.5 + 1
B = 5.(42003 + 42002 + ... + 4) + 1 chia 5 dư 1
=> B = 5.k + 1 (k là số chia hết cho 4)
=> A = 75.(5.k + 1) + 25
=> A = 75.5k + 75 + 25
=> A = (...00) + 100
=> A = (...00) chia hết cho 100 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
A= 75. (42004+.......+4+1) + 25
= 25 . (4-1) . (42004+.....+4+1) + 25
= 25.[4.(42004+......+4+1) - (42004+......+4+1)] + 25
= 25.[ (4+ 42+........+ 42005 ) - ( 1+ 4 +........+42004)] + 25
= 25.(42005 - 1) + 25
= 25. 42005- 25 +25
= 25. 42005
= (25. 4). 42004
= 100. 22004
Mà 100 chia hết cho 100 => 100. 22004 chia hết cho 100
=> A chia hết cho 100 ( đccm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=75\left(4^{2004}+...+4+1\right)+25\)
\(=25\left(4-1\right)\left(4^{2004}+...+4+1\right)+25\)
\(=25\left[4\left(4^{2004}+...+4+1\right)-\left(4^{2004}+...+4+1\right)\right]+25\)
\(=25\left[\left(4+4^2+...+4^{2005}\right)-\left(1+4+...+4^{2004}\right)\right]+25\)
\(=25\left(4^{2005}-1\right)+25\)
\(=25.4^{2005}-25+25\)
\(=100.4^{2004}⋮100\)
![](https://rs.olm.vn/images/avt/0.png?1311)
dat A=75*(4^2004+4^2003+...+4^2+4+1)+25
B=4^2004+4^2003+...+4^2+4+1
4B=4+4^2+4^3+...+4^2005
3B=4^2005-1
B=(4^2005-1)/3
A=75*(4^2005-1)/3+25
A=25*(4^2005-1)+25
A=25*4*4^2004-25+25
A=100*4^2004
Vay A chia het cho 100
k cho minh nhieu nha
khong can biet ơi cứu tớ