Q1.3MCQ1 mark, −0.25The solution to the recurrence equation T(2k)=3T(2k−1)+1,T(1)=1T(2^k) = 3T(2^{k-1})+1, T(1) =1T(2k)=3T(2k−1)+1,T(1)=1 is(A)2k2^k2k(B)(3k+1−1)2\frac{(3^{k+1}-1)}{2}2(3k+1−1)(C)3log2k3^{\log_2 k}3log2k(D)2log3k2^{\log_3 k}2log3kSign in to save attempts.Show answer← PreviousNext →SolutionsDiscussionLoading…Sign in with Google