Question

O(3^n), O(n^2),O(1), O(n log n) increasing order

Answer 1

\(O(1)\) is evidently the smallest

Answer 2

O(3^n) is the biggest its obvious

Answer 3

then \(\log(n)\) grows slower than \(n\) so \(O(n\log(n))\) is smaller than \(O(n^2)\)

Answer 4

simpler method, put n as very large number, then arrange your answer in increasing order.
that would be the order of O's

Answer 5

thanks to all......
O(1)<O(n long n) < O(n^2)<O(3^n)
Am I right???/

Answer 6

yup :)
good work.

Answer 7

yes

Answer 8

prove mathematical induction (sum 2^0 + 2^1 + ...+ 2^n is 2^n+1 - 1 for all n>=1

Answer 9

how to prove???