SOLUTION: find the general terms of the following sequences a. 4,7,12,19,28 b. 3,-5,7,-9,11 c. 2,6,12,20,30 d. 3,-6,12,-24,48

a.    4, 7, 12, 19, 28

Subtract term by term the sequence of perfect squares:

#1:   1, 4,  9, 16, 25

whose general term is n²

and we get the sequence
 
#2:   3, 3,  3,  3,  3

whose general term is just 3

So the general term of the given sequence is
gotten by reversing the operations we just
did with the sequences, with their general
terms.  We take the general term of the
the last sequence #2: which is 3, and since
we subtracted sequence #1 from the given
sequence, we will do the opposite and add 
the general term for #1, which is n², so
the general term of the given sequence (a)
is  

an = 3 + n²  

--------------------------------------------

b.    3,-5,7,-9,11

The signs alternate, so

we divide term by term the sequence:

#1:   1,-1,1,-1,1

whose general term is (-1)n+1

and get the sequence of positive terms

#2:   3, 5,7, 9, 11

We subtract the sequence

#3:   1, 1,1, 1,  1, whose general term is just 1

and we get the sequence of even integers:

#4:   2, 4,6, 8, 10, whose general term is 2n

So the general term of the given sequence is
gotten by reversing the operations we just
did with the sequences, with their general
terms.  

We take the general term of the
the last sequence #4: which is 2n, and since
we subtracted sequence #3 from sequence #2,
we will do the opposite with the general term
of #3 which is 1 to 2n, and now we have the
general term for #2, which is 2n+1. 

Since we divided the given sequence (b) by
sequence #1, we do the opposite and multiply
2n+1 by the general term of sequence #1 which
is  (-1)n+1, and get the general
term for (b)

an = (-1)n+1(2n+1) 

-------------------------------------------

c.    2 ,6, 12, 20, 30

Subtract term by term the sequence of perfect squares:

#1:   1, 4,  9, 16, 25

whose general term is n²

and we get the sequence
 
#2:   1 ,2,  3,  4,  5

whose general term is just n

So the general term of the given sequence is
gotten by reversing the operation we just
did with the sequences, with their general
terms.  We take the general term of the
the last sequence #2: which is n, and since
we subtracted sequence #1 from the given
sequence, we will do the opposite and add 
the general term for #2, which is n², so
the general term of the given sequence is  

an = n + n²

or if you like, you can factor out n:

an = n(1 + n)  

-------------------------

d.    3,-6,12,-24,48

The signs alternate, so

we divide term by term the sequence:

#1:   1,-1, 1, -1, 1

whose general term is (-1)n+1

and get the sequence of positive terms

#2:   3, 6,12, 24, 48

We divide each term by the sequence of 3's

#3:   3, 3, 3,  3,  3

whose general term is just 3, and we get

#4:    1, 2, 4, 8, 16 

which is the sequence of powers
of 2, starting with the 0 power of 2  

#4:    20,21,22,23, 24

And the sequence of the exponents is

#5:    0, 1, 2, 3,  4

and if we add the sequence of 1's

#6:    1, 1, 1, 1,  1

whose general term is just 1, we get

#7:    1, 2, 3, 4,  5

whose general term is just n.

So the general term of the given sequence is
gotten by reversing the operation we just
did with the sequences, with their general
terms.  

We take the general term of the
the last sequence #7: which is n, and since
we added sequence #6 to sequence #5 to get 
sequence #7, we will do the opposite and 
subtract the general term for #6, which is 1, 
so the general term of sequence #5
is n-1.  

#5 was the sequence of exponents in #4, so the
general term for sequence #4 is 2n-1.

Since we divided sequence #2 by the sequence #3 
of 3's to get #4, we will do the opposite and 
multiply by #3's general term, 3, and so the 
general term for sequence #2 is 3·2n-1.

Since we divided the given sequence (c) by
sequence #1 to get sequence #2, we do the opposite
and multiply 3·2n-1 by the general term of
sequence #1 which is  (-1)n+1, and
get the general term for (c):   

an = (-1)n+1·3·2n-1.

or we might want to put the 3 in front:

an = 3(-1)n+12n-1.

Edwin