What we know from the given information:
__ __ __ __ __ __ 35 terms 1 through 7
1 3 5 ___ ___ ___ first differences
What we can determine, since the first differences for a quadratic sequence form an arithmetic sequence:
__ __ __ __ __ __ 35 terms 1 through 7
1 3 5 7 9 11 first differences
And now we can work backwards and up in the array to find the terms of the sequence:
-1 0 3 8 15 24 35 terms 1 through 7
1 3 5 7 9 11 first differences
Now we want to find the quadratic expression for the n-th term of the sequence. Here is the standard method....
The expression is of the form 
.
Use the first three terms of the sequence (n=1, 2, and 3) to get three equations in a, b, and c and solve that system of equations.
n=1: 
n=2: 
n=3: 
The formula for the n-th term of the sequence is 
A student much more familiar with this kind of problem might use a different method for finding the expression for the n-th term, using the following basic principle:
*****************************************************
If the k-th differences are the constant d,
then the polynomial is of degree k,
and the leading coefficient is 
*****************************************************
In this problem, the constant 2nd difference is 2, so the polynomial is degree 2, and the leading coefficient is 
So we know the quadratic function is of the form
Now we have only two unknowns, so we only need two equations.
n=1: 1+b+c = -1; b+c=-2
n=2: 4+2b+c = 0; 2b+c=-4
b=-2
c=0
And again (of course) we end up with the same formula for the n-th term of the sequence:
