$\frac{1}{1-2x^{2}}$ = $\frac{1}{1-\sqrt{2}x}$ * $\frac{1}{1+\sqrt{2}x}$ ................(1)
Now, we know that,
$\frac{1}{1-ax}$ = 1+ ax + a^{2}x^{2 } + a^{3}x^{3 } +.................. ..........(2)
$\frac{1}{1+ax}$ = 1- ax + a^{2}x^{2 } - a^{3}x^{3 } +.................. ...........(3)
Now, from (1) , we observe that the result is the product of 2 functions.
If the coefficients of function 1 are a_{0}, a_{1}, a_{2................}
and that of function 2 are b_{0}, b_{1}, b_{2................}
then the result of product of two functions is a function with terms having coefficients r_{k} corresponding to ''k'th term of the sequence. The coefficients can be calculated as follows:
r_{k} = a_{k}b_{0} + a_{k-1}b_{1} + a_{k-2}b_{2.........................}+ a_{0}b_{k}
From (2) and (3), we get,
r_{0} = 1
r_{1} = 0
r_{2} = a^{2}
r_{3} = 0
........
We observe that when 'k' is odd, r_{k} = 0. If 'k' is even, r_{k} = a^{k}
Thus, the terms of the sequence are {1, 0, a^{2}, 0, a^{4}, 0, ........................}
Now, substitute $\sqrt{2}$ for 'a' to get the actual sequence.