As you observed, the nth term is the sum of the first n squares. 

1 = 1² 

5 = 1²+2² 

14 = 1²+2²+3² 

30 = 1²+2²+3²+4²

So the nth term is 1²+2²+3²+...+n²

That should be related to the sum 1+2+3+...+n

1+2+3+...+n is the sum of an arithmetic series with a1=1 and d=1

Using the formula for the sum of the an arithmetic series to n terms:

1+2+3+...+n =  or 

Let's divide each term of our sequence by the sum of the 
first n natural numbers, and see if we get a recognizable
pattern:

1/(1) = 1

5/(1+2) = 5/3

14/(1+2+3) = 14/6 = 7/3

30/(1+2+3+4) = 30/10 = 3

Now if we write 1 as 3/3 and 3 and 9/3 we do have a 
recognizable pattern

3/3, 5/3, 7/3, 9/3, ...

The numerators go 3,5,7,9,... each of which is 1 more than 2,4,6,8,...
which has nth term 2n

So 3,5,7,9, has nth term 2n+1

So 3/3, 5/3, 7/3, 9/3 has nth term (2n+1)/3

Since we got that sequence by DIVIDING our sequence by the sums of the 
first n natural numbers, the nth term of our sequence is gotten by 
MULTIPLYING those two nth terms.  So the nth ordered pair is given
by the equation:
 


or



Edwin