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Let Tn be the nth triangle number
n Tn 8Tn +1
1 1 9
2 3 25
3 6 49
4 10 81
5 15 121

We can see that each term is a square number. They are actually the squares of (2n+1)

Proof:
Tn x 2 can be written as:
(n + 1) + (n-1 + 2) + ... (2 + n-1) + (1 + n) =
(n+1) + (n+1)+... (n+1) + (n+1)
We now have n lots of (n+1)
Therefore Tn x 2 = n(n+1)
Tn = (n^2+n)/2
8Tn = 4n^2+4n
8Tn + 1 = 4n^2+4n+1
= (2n+1)^2