To find the next two terms of the sequence, we need to first find the common difference between consecutive terms:
52 - 44 = 8
64 - 52 = 12
80 - 64 = 16
We notice that the common difference is increasing by 4 for each term. Therefore, the next two terms of the sequence are:
80 + 20 = 100
100 + 24 = 124
To determine the nth term of the quadratic sequence, we can use the formula:
an = a1 + (n-1)d + bn^2
where a1 is the first term, d is the common difference, b is the coefficient of n^2, and n is the term number.
Using the first four terms of the sequence, we can form a system of equations:
44 = a1 + b
52 = a1 + d + b
64 = a1 + 2d + b
80 = a1 + 3d + b
Solving for a1 and b, we get:
a1 = 20
b = 24
Substituting these values into the formula for an, we get:
an = 20 + (n-1)4 + 24n^2
an = 24n^2 + 4n - 4
To find the 30th term of the sequence, we simply substitute n = 30 into the formula we just derived:
a30 = 24(30)^2 + 4(30) - 4
a30 = 21,596
To prove that the quadratic sequence will always have even terms, we notice that the first term is even (44 = 2 x 22), and the common difference is even (8 = 2 x 4). Therefore, every term of the sequence can be expressed as an even number plus an even multiple of n^2, which is always even. Hence, the quadratic sequence will always have even terms.