4 a_{n} - a_{n-2} = 0 ...............from the given recurrance
Now, adding and subtracting a_{n-3},
\therefore 4 a_{n} - a_{n-2} + a_{n-3 }- a_{n-3} = 0
\therefore 4 a_{n} - a_{n-2} + a_{n-3 }- 4 a_{n-1} = 0 (from the recurrance for a_{n-1})
\therefore 4 a_{n} _{ }- 4 a_{n-1} - a_{n-2} + a_{n-3}= 0
This is our normal recurrance now.
\therefore 4t^{3} - 4t^{2} - t +1 =0
By trial and error, 1/2 is one of the factors.
Now, we perform synthetic division.
1/2 | 4 -4 -1 1
| +2 -1 -1
---------------------------------------
4 -2 -2 0
So, factorisation is
(t-1/2)(t -1)( 2t +1) = 0
So, factors are t= 1/2 , -1/2, 1
So, f(n) = c1 . (1/2)^{n} + c2. (1)^{n} + c3. (-1/2)^{n}
Now, putting the values of a0 and a1, we get answer B.