GATE | GATE-CS-2002 | Question 3

The solution to the recurrence equation T(2^k) = 3 T(2^k-1) + 1, T (1) = 1, is:
(A) 2^k
(B) (3^{k + 1} – 1)/2
(C) 3^log2k
(D) 2^log3k


Answer: (B)

Explanation: We have

T (2^k) = 3  T (2^k-1) + 1

= 3²  T (2^k-2) + 1 + 3
= 3³  T (2^k-3) + 1 + 3 + 9
. . . (k steps of recursion (recursion depth))
= 3^k  T (2^k-k) + (1 + 3 + 9 + 27 + … + 3^k-1)
= 3^k + ( ( 3^k – 1 ) / 2 )
= ( (2 * 3^k) + 3^k – 1 )/2
= ( (3 * 3^k) – 1 ) / 2
= (3^k+1 – 1) / 2

Hence, B is the correct choice.

Please comment below if you find anything wrong in the above post.

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