$\frac{1}{1-2x^{2}}$ = $\frac{1}{1-\sqrt{2}x}$ * $\frac{1}{1+\sqrt{2}x}$ ................(1)
Now, we know that,
$\frac{1}{1-ax}$ = 1+ ax + a2x2 + a3x3 +.................. ..........(2)
$\frac{1}{1+ax}$ = 1- ax + a2x2 - a3x3 +.................. ...........(3)
Now, from (1) , we observe that the result is the product of 2 functions.
If the coefficients of function 1 are a0, a1, a2................
and that of function 2 are b0, b1, b2................
then the result of product of two functions is a function with terms having coefficients rk corresponding to ''k'th term of the sequence. The coefficients can be calculated as follows:
rk = akb0 + ak-1b1 + ak-2b2.........................+ a0bk
From (2) and (3), we get,
r0 = 1
r1 = 0
r2 = a2
r3 = 0
........
We observe that when 'k' is odd, rk = 0. If 'k' is even, rk = ak
Thus, the terms of the sequence are {1, 0, a2, 0, a4, 0, ........................}
Now, substitute $\sqrt{2}$ for 'a' to get the actual sequence.