RÄspuns :
a) n-1|n+1
n-1| n-1
n+1+n-1=2 deci n-1 | 2 ān-1 ā D2
n-1ā{-2,-1,1,2} nā{-1,0,2,3} nāNāS:nā{0,2,3}
b)n-1 | 3n+1ān-1| 3n+1
n-1 | n-1 ān-1| 3n-3
3n+1-3n+3=4 deci n-1| 4 ān-1 ā D4 n-1ā {-4,-2,-1,1,2,4}
nā{-3,-1,0,2,3,5}
nāNāS:nā{0,2,3,5}
c)2n-1|3n-1 ā2n-1 |6n-2
2n-1|2n-1 ā2n-1 |6n-3
6n-2-6n+3=1 deci 2n-1 | 1ā2n-1 āD1 2n-1ā {-1,1} 2nā{0,2}
nā{0,1} nāNāS:nā{0,1}
2.S=3 + 3Ā² + 3Ā³ + ... + 3Ā¹Ā²
avem 12 termeni putem sa ii grupa cate 3
S=(3 + 3Ā² + 3Ā³) + ...(3Ā¹ā°+3Ā¹Ā¹ + 3Ā¹Ā²)=
=3(1 + 3 + 3Ā²) + ...3Ā¹ā°(1+3 + 3Ā²)=
=3(1 + 3 + 9) + ...3Ā¹ā°(1+3 + 9)=
=13(3+...+3Ā¹ā°) deci divizibil cu 13
avem 12 termeni putem sa ii grupa cate 4
S=3 + 3Ā² + 3Ā³ + ... + 3Ā¹Ā²=(3 + 3Ā² + 3Ā³ +3ā“+ ...3ā¹+3Ā¹ā°+3Ā¹Ā¹ + 3Ā¹Ā²)=
=3(1+3+3Ā²+3Ā³)+...+3ā¹(1+3+3Ā²+3Ā³)=
=3(1+3+9+27)+...+3ā¹(1+3+9+27)=
=3Ć40+...+3ā¹Ć40=
=5(3Ć8+...+3ā¹Ć8 deci divizibil cu 5
n-1| n-1
n+1+n-1=2 deci n-1 | 2 ān-1 ā D2
n-1ā{-2,-1,1,2} nā{-1,0,2,3} nāNāS:nā{0,2,3}
b)n-1 | 3n+1ān-1| 3n+1
n-1 | n-1 ān-1| 3n-3
3n+1-3n+3=4 deci n-1| 4 ān-1 ā D4 n-1ā {-4,-2,-1,1,2,4}
nā{-3,-1,0,2,3,5}
nāNāS:nā{0,2,3,5}
c)2n-1|3n-1 ā2n-1 |6n-2
2n-1|2n-1 ā2n-1 |6n-3
6n-2-6n+3=1 deci 2n-1 | 1ā2n-1 āD1 2n-1ā {-1,1} 2nā{0,2}
nā{0,1} nāNāS:nā{0,1}
2.S=3 + 3Ā² + 3Ā³ + ... + 3Ā¹Ā²
avem 12 termeni putem sa ii grupa cate 3
S=(3 + 3Ā² + 3Ā³) + ...(3Ā¹ā°+3Ā¹Ā¹ + 3Ā¹Ā²)=
=3(1 + 3 + 3Ā²) + ...3Ā¹ā°(1+3 + 3Ā²)=
=3(1 + 3 + 9) + ...3Ā¹ā°(1+3 + 9)=
=13(3+...+3Ā¹ā°) deci divizibil cu 13
avem 12 termeni putem sa ii grupa cate 4
S=3 + 3Ā² + 3Ā³ + ... + 3Ā¹Ā²=(3 + 3Ā² + 3Ā³ +3ā“+ ...3ā¹+3Ā¹ā°+3Ā¹Ā¹ + 3Ā¹Ā²)=
=3(1+3+3Ā²+3Ā³)+...+3ā¹(1+3+3Ā²+3Ā³)=
=3(1+3+9+27)+...+3ā¹(1+3+9+27)=
=3Ć40+...+3ā¹Ć40=
=5(3Ć8+...+3ā¹Ć8 deci divizibil cu 5