Chứng tỏ rằng 5^1+5^2+5^3+...+5^2016chia hết cho31
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2+22+23+....+22016 =(2+22)+ (23+24 )+....+(22015 +22016)
=2*(1+2)+.....+22015 *(1+2)
=2*3+.......+22015*3
=3(2+.......+22015) chia hết cho 3
2 + 2^2 +2^3 + ...... +2^2016
= (2+2^2) + .. +(2^2015 + 2^2016)
= 2.3 + .. + 2^2015.3
= 3.(2+2^3+...+2^2015)
Chia hết cho 3
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a)\(H=1+5+...+5^{120}\)
\(=\left(1+5\right)+...+\left(5^{119}+5^{120}\right)\)
\(=1\cdot\left(1+5\right)+...+5^{119}\left(1+5\right)\)
\(=1\cdot6+...+5^{119}\cdot6\)
\(=6\cdot\left(1+...+5^{119}\right)⋮6\left(DPCM\right)\)
b)\(H=1+5+...+5^{120}\)
\(=\left(1+5+5^2\right)+...+\left(5^{118}+5^{119}+5^{120}\right)\)
\(=1\left(1+5+5^2\right)+...+5^{118}\left(1+5+5^2\right)\)
\(=1\cdot31+...+5^{118}\cdot31\)
\(=31\cdot\left(1+...+5^{118}\right)⋮31\left(DPCM\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(A=5+5^2+..+5^{12}\)
\(5A=5\cdot\left(5+5^2+..5^{12}\right)\)
\(5A=5^2+5^3+...+5^{13}\)
\(5A-A=\left(5^2+5^3+...+5^{13}\right)-\left(5+5^2+...+5^{12}\right)\)
\(4A=5^2+5^3+...+5^{13}-5-5^2-...-5^{12}\)
\(4A=5^{13}-5\)
\(A=\dfrac{5^{13}-5}{4}\)
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\(\text{Đặt A=}1+5+5^2+5^3+...+5^{403}+5^{404}\)
\(=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\)
\(=\left(1+5+25\right)+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\)
\(=31+5^3.31+...+5^{402}.31\)
\(=31.\left(1+5^3+...+5^{402}\right)\text{chia hết cho 31}\)
=> A chia hết cho 31 => đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ghép các số lại
1+5+5^2=31
5^3+5^4+5^5=5^3.(1+5+5^2)=5^3.31
Dễ r đung ko?