+1 vote
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The solution of the recursive relation
for n ≥ 2, a0 = 1, a1 = 0

1.
2.

3.
4.

Put n = 1 we get $\left ( \frac{1}{2} \right )^{3} - \left ( \frac{-1}{2} \right )^{3} = \left ( \frac{1}{8} \right ) + \left ( \frac{1}{8} \right ) = \left ( \frac{1}{4} \right )$ which is matched with given reccurence relation .= 1/4

Put n= 2 we get $\left ( \frac{1}{2} \right )^{4} - \left ( \frac{-1}{2} \right )^{4} = \left ( \frac{1}{16} \right ) - \left ( \frac{1}{16} \right ) = \left ( 0 \right )$  which is matched with given reccurence relation .= 0

Put n= 3 we get $\left ( \frac{1}{2} \right )^{5} - \left ( \frac{-1}{2} \right )^{5} = \left ( \frac{1}{32} \right ) + \left ( \frac{1}{32} \right ) = \left ( \frac{1}{16} \right )$  which is matched with given reccurence relation .= (1/4 )/4 = (1/16)

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by putting n=2 ,we will get 1/4 as a0 =1

and answer given is B actually
now check always try to put in these quetion :) but no wrongly like i did :)
thanks a lot @anirudh..
+1 vote

4 an - an-2 = 0 ...............from the given recurrance

\therefore  4 an - an-2 + an-3 - an-3 = 0

\therefore  4 an - an-2 + an-3 4 an-1 = 0      (from the recurrance for an-1)

\therefore   4 an 4 an-1 - an-2 + an-3= 0

This is  our normal recurrance now.

\therefore 4t3 - 4t2 - 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.