edited by
755 views
0 votes
0 votes

The recurrence equation:

T(1) = 1

T(n) = 2T(n - 1) + n, n ≥ 2

evaluates to

(a) 2n + 1 - n – 2                                                            (b) 2n – n

(c) 2n + 1 – 2n – 2                                                         (d) 2n – n

Solution: Option (a)

edited by

1 Answer

Related questions

1 votes
1 votes
1 answer
1
1 votes
1 votes
1 answer
2
iarnav asked Jul 29, 2017
2,438 views
Given RR as -T(n) = 2T(n/2)+n ; n>1T(1) = 1Solve this using only BACK SUBSTITUTION method? Note - I am stuck at T(n)= 2^k.T(n/2^k)+(2^k-1).nand I'm putting 2^k=n Please h...
2 votes
2 votes
1 answer
3
indrajeet asked Feb 10, 2016
519 views
T(n) = 2*T(n/2) + n*logn please explain
4 votes
4 votes
1 answer
4
Lakshman Bhaiya asked Jan 15, 2018
3,248 views
Which of the following represents most appropriate asymptotic solution for given reccurance:(A) O(n)(B) O(log n)(C) O(log log n)(D) O(log n)2