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Đặt \(A=1+5+5^2+5^3+...+5^{402}+5^{403}+5^{404}\)
\(\Rightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{399}+5^{400}+5^{401}\right)+\left(5^{402}+5^{403}+5^{404}\right)\)
\(\Rightarrow A=31.1+31.5^3+...+31.5^{402}\)
\(\Rightarrow A=31\left(1+5^3+5^6+...+5^{402}\right)\)
\(\Rightarrow A⋮31\left(đpcm\right)\)
\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)
\(B=5+5^2+5^3+...+5^{88}+5^{89}+5^{90}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{88}+5^{89}+5^{90}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{88}\left(1+5+5^2\right)\)
\(=31\left(5+5^4+...+5^{88}\right)⋮31\)
a, Ta có:
2 + 2 2 + 2 3 + 2 4 + . . . + 2 99 + 2 100
= 2 + 2 2 + 2 3 + 2 4 + 2 5 +...+ 2 96 + 2 97 + 2 98 + 2 99 + 2 100
= 2. 1 + 2 + 2 2 + 2 3 + 2 4 +...+ 2 96 1 + 2 + 2 2 + 2 3 + 2 4
= 2 . 31 + 2 6 . 31 + . . . + 2 96 . 31
= 2 + 2 6 + . . . + 2 96 . 31 chia hết cho 31
b, Ta có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 1 + 5 + 5 3 1 + 5 + 5 5 1 + 5 + . . . + 5 149 1 + 5
= 5 . 6 + 5 3 . 6 + 5 5 . 6 + . . . + 5 149 . 6
= ( 5 + 5 3 + 5 5 + . . . + 5 149 ) . 6 chia hết cho 6
Ta lại có:
5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150
= 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 +...+ 5 145 + 5 146 + 5 147 + 5 148 + 5 149 + 5 150 (có đúng 25 nhóm)
= [ ( 5 + 5 4 ) + ( 5 2 + 5 5 ) + ( 5 3 + 5 6 ) ] + ... + [ 5 145 + 5 148 ) + ( 5 146 + 5 149 ) + ( 5 147 + 5 150 ]
= [ 5 ( 1 + 5 3 ) + 5 2 ( 1 + 5 3 ) + 5 3 ( 1 + 5 3 ) ] + ... + [ 5 145 1 + 5 3 ) + 5 146 ( 1 + 5 3 ) + 5 147 ( 1 + 5 3 ]
= ( 5 . 126 + 5 2 . 126 + 5 3 . 126 ) + ... + ( 5 145 . 126 + 5 146 . 126 + 5 147 . 126 )
= ( 5 + 5 2 + 5 3 ) . 126 + ( 5 7 + 5 8 + 5 9 ) . 126 + ... + ( 5 145 + 5 146 + 5 147 ) . 126
= 126.[ ( 5 + 5 2 + 5 3 ) + ( 5 7 + 5 8 + 5 9 ) + ... + ( 5 145 + 5 146 + 5 147 ) ] chia hết cho 126.
Vậy 5 + 5 2 + 5 3 + 5 4 + 5 5 + 5 6 + . . . + 5 149 + 5 150 vừa chia hết cho 6, vừa chia hết cho 126
\(B=\left(1+5+5^2\right)+...+5^6\left(1+5+5^2\right)=31\left(1+...+5^6\right)⋮31\)
b, \(B=5+5^2+5^3+5^4+...+5^{11}+5^{12}\)
\(B=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{11}+5^{12}\right)\)
\(B=30+5^2\left(5+5^2\right)+...+5^{10}\left(5+5^2\right)\)
\(B=30+5^2\cdot30+...+5^{10}\cdot30\)
\(B=\left(1+5^2+...+5^{10}\right)\cdot30\)\(⋮30\)
+) \(B=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{10}+5^{11}+5^{12}\right)\)
\(B=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{10}\left(1+5+5^2\right)\)
\(B=5\cdot31+5^4\cdot31+...+5^{10}\cdot31\)
\(B=\left(5+5^4+...+5^{10}\right)\cdot31\)\(⋮31\)
\(1+5+5^2+...+5^{404}\)
\(=5^3\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{404}\left(1+5+5^2\right)\)
\(=\left(1+5+5^2\right)\left(5^3+5^4+...+5^{403}+5^{404}\right)\)
\(=31.\left(5^3+5^4+...+5^{403}+5^{404}\right)\)
Vậy tổng trên chia hết cho 31
1 + 5 + 52 + .... + 5404
= ( 1 + 5 ) + ( 52 + 53 ) + ... + ( 5403 + 5404 )
= 6 + 52 . ( 1 + 5 ) + ... + 5403 . ( 1 + 5 )
=6 + 52 . 6 + ... + 5403 . 6
= 6 . ( 1 + 52 + ... + 5403 )
= 3 . 2 . ( 1 + 52 + .... + 5403 ) chia hét cho 3