How to complete these Maths equations

1) given Un = an^2 + bn , U4 = 28 U8 = 120 find the values of a and b.

2)show by completing the square or otherwise, that all the terms in the sequence given by: Un = n^2 - 8n + 19 are positive.

3) what is the smallest term in the sequence generated by: Un = n^2 - 6n + 13?

4) a) if Un = 2n - 3 show that the terms of the sequence given by U2n + 1 are odd no.s

b) given that Un = n^2 - 2n - 5 show that U3n + 1 generates a sequence in which all terms are multiples of three. Hence write down the first three terms in the sequence.

Any help would be really appreciated been doing Maths for an hour and can't seem to get these. Thnx xxxx

1) U4 = 16a + 4b
so 16a + 4b = 28 ... eqn 1
U8 = 64a + 8b
so 64a +8b = 120 ... eqn 2
Multiplying eqn 1 by 2, we get
32a + 8b = 56 ... eqn 3
Subtracting eqn 3 from eqn 2, we get
32a = 64
a = 2
Substituting into eqn 1, we get
32 + 4b = 28
4b = - 4
b = - 1

2) (completing the square)
Un = n^2 - 8n + 19
= n^2 - 8n + 16 + 3
= (n - 4)^2 + 3
(n - 4)^2 cannot be less than 0
so (n - 4)^2 + 3 cannot be less than 3
So all terms given by Un = n^2 - 8n + 19 are positive

3) Un = n^2 - 6n + 13
= n^2 - 6n + 9 + 4
= (n - 3)^2 + 4
The smallest possible value of (n - 3)^2 is 0
(when n = 3)
So the smallest term in the sequence is 0 + 4 = 4.

4a) Un = 2n - 3
so U2n + 1 = 2(2n + 1) - 3
which must always be odd as it is an even number minus an odd number.

4b) Un = n^2 - 2n - 5
so U3n + 1 = (3n + 1)^2 - 2(3n + 1) - 5
= 9n^2 + 6n + 1 - 6n - 2 - 5
= 9n^2 - 6
= 3(3n^2 - 2)
Hence all terms of U3n + 1 are multiples of 3.
The first 3 terms are 3, 30 and 75.