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## relation to Non-Recursive formulas for K(n) and K(2n)

A known recursive formula for square triangular numbers (of the form T_k= k/2*{k+1}) is K_n = 34K_{n-1}-K_{n-2}+2.

So knowing the first two square triangular numbers i.e. K_1 = T_1 =1 and K_2= T_8 = 36, all other successive square triangular numbers can be obtained , e.g.

K_3 = 34 * K_2 – K_1 + 2 = 34 * 36 -1 + 2 = 1225

K_4 = 34 * K_3 – K_2 + 2 = 34 * 1225 - 36 + 2 = 41616

The following non- recursive formula also gives nth Square Triangular number in terms of variable n.

K_n = ({1 +sqrt[2]}^2n – {1-sqrt[2]}^2n)^2 / (4sqrt[2])^2

Now from inspection it appears that given the square triangular number K_n, the square triangular number K_2n =T_(8*K_n), but I have difficulty using the non recursive formula for K_n and K_2n to verify this. Is anyone willing to work this out for me?
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