Chứng minh 1+5^1+5^2+5^3+5^4+.....+5^101 chia hết cho 6
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1) \(B=1+5+5^2+5^3+....+5^{101}\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+.....+\left(5^{100}+5^{101}\right)\)
\(=\left(1+5\right)+5^2\left(1+5\right)+....+5^{100}\left(1+5\right)\)
\(=\left(1+5\right)\left(1+5^2+....+5^{100}\right)\)
\(=6\left(1+5^2+...+5^{100}\right)\)\(⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình giúp cho đáp án đúng 100%
5^2003+5^2002+5^2001 chia hết cho 31
=5^2001.(1+5+5^2)
=5^2001.31 chia hết cho 3
hai bài kia tương tự rất dễ đúng ko
Ta có: 52003 + 52002 + 52001
= 52001.(1 + 5 + 25)
= 52001 . 31 chia hết cho 31
Ta có: 1 + 7 + 72 + ...... + 7101
= (1 + 7) + (72 + 73) + ..... + (7100 + 7101)
= 1.8 + 72.(1 + 7) + ..... + 7100.(1 + 7)
= 1.8 + 72.8 + ..... + 7100 . 8
= 8.(1 + 72 + ..... + 7100) chia hết cho 8
![](https://rs.olm.vn/images/avt/0.png?1311)
1 +5+ 52 +53 + ...+ 5100 + 5101
= (1 + 5) + (52 + 53) + ... + (5100 + 5101)
= 6 + 52(1 + 5) + ... + 5100.(1 + 5)
= 6 + 52.6 + ... + 5100.6
= 6.(1 + 52 + ... + 5100) \(⋮\)6
\(1+5+5^2+.....+5^{101}⋮6\)
\(=\left(1+5\right)+\left(5^2+5^3\right)+.....+\left(5^{100}+5^{101}\right)\)
\(=6+\left(5^2.1+5^2.5\right)+.....+\left(5^{100}.1+5^{100}.5\right)\)
\(=6+5^2.\left(1+5\right)+.....+5^{100}.\left(1+5\right)\)
\(=6+5^2.6+....+5^{100}.6\)
\(=\left(1+5^2+....+5^{100}\right).6⋮6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 52003 + 52002 + 52001
= 52001.(52 + 5 + 1)
= 52001 . 31 chia hết cho 31
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
( 21 + 22 ) + ( 23 + 24 ) + ... + ( 22009 + 22010 )
= 2. ( 1 + 2 ) + 23 . ( 1 + 2 ) + ... + 22009 . ( 1 + 2 )
= 3 . ( 2 + 23 + ... + 22009 ) chia hết cho 3. => ĐPCM
1+5^1+5^2+5^3+5^4+.....+5^101
= ( 1+ 5^1) + (5^2 + 5^3) + ...+(5^100 + 5^101)
= 1.(1+5) + 5^2. (1+5)+...+5^100.(1+5)
=(1+5^2+...+5^100).6 chia hết cho 6(vì 6 chia hết cho 6 )
vậy 1+5^1+5^2+5^3+5^4+.....+5^101 chia hết cho 6
Gọi dãy trên là A
A=(1+5^1)+(5^2+5^3)+...+(5^100+5^101)
A=1.(1+5^1)+5^2.(1+5^1)+...+5^100.(1+5^1)
A=1.6+5^2.6+...+5^100.6
A=6.(1+5^2+...+5^100) chia hết cho 6