Let a (n) denote the n-th term of the given sequence.
Check the forward differences, and denote the n-th difference by b (n). That is,
b (n) = a (n + 1) - a (n)
These so-called first differences are
b (1) = a (2) - a (1) = 25 - 7 = 18
b (2) = a (3) - a (2) = 51 - 25 = 26
b (3) = a (4) - a (3) = 85 - 51 = 34
b (4) = a (5) - a (4) = 127 - 85 = 42
Now consider this sequence of differences,
18, 26, 34, 42, …
and notice that the difference between consecutive terms in this sequence b is 8:
26 - 18 = 8
34 - 26 = 8
42 - 34 = 8
and so on. This means b is an arithmetic sequence, and in particular follows the rule
b (n) = 18 + 8 (n - 1) = 8n + 10
for n ≥ 1.
So we have
a (n + 1) - a (n) = 8n + 10
or, replacing n + 1 with n,
a (n) = a (n - 1) + 8 (n - 1) + 10
a (n) = a (n - 1) + 8n + 2
We can solve for a (n) by iteratively substituting:
a (n) = [a (n - 2) + 8 (n - 1) + 2] + 8n + 2
a (n) = a (n - 2) + 8 (n + (n - 1)) + 2×2
a (n) = [a (n - 3) + 8 (n - 2) + 2] + 8 (n + (n - 1)) + 2×2
a (n) = a (n - 3) + 8 (n + (n - 1) + (n - 2)) + 3×2
and so on. The pattern should be clear; we end up with
a (n) = a (1) + 8 (n + (n - 1) + … + 3 + 2) + (n - 1)×2
The middle group is the sum,

so that
a (n) = a (1) + (4n ² + 4n - 8) + 2 (n - 1)
a (n) = 4n ² + 6n - 3