13 13 votes Consider the following sequence: $s_1 = s_2 = 1$ and $s_i = 1 + \min \left({s_{i-1}, s_{i-2}}\right) \text{ for } i > 2$. Prove by induction on $n$ that $s_n=⌈\frac{n}{2}⌉$. Set Theory & Algebra gatecse-2000 set-theory&algebra mathematical-induction descriptive + – Kathleen 2.5k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 27 27 votes $s_3 = 1 + \min(s_1, s_2) \\= 1 + \min(1, 1) = 2 = \lceil \frac{3}{2} \rceil$. So, base condition of induction satisfied. Assume, $s_{n-2} =\lceil \frac{n-2}{2} \rceil$ and $s_{n-1} =\lceil \frac{n-1}{2} \rceil$ (Induction hypothesis) Now, we have to prove, $s_n = \lceil \frac{n}{2} \rceil$ $s_n = 1 + \min(s_{n-1}, s_{n-2}) \\= 1 + \lceil \frac{n-2}{2} \rceil \\= 1 + \lceil \frac{n}{2} \rceil -1 \\=\lceil \frac{n}{2}\rceil$ (Hence, proved) Arjun answered Aug 12, 2015 • selected Dec 1, 2015 by Akash Kanase Arjun comment Share Follow See all 2 Comments 2 2 Comments reply Warrior commented Aug 6, 2017 reply Follow flag Nice explanation.Thank you sir. :) 0 0 replyShare Kiyoshi commented Dec 23, 2021 reply Follow flag Sequence is 1 1 2 2 3 3 4 4 5 5 6 6 7 7 ……… Sn is representing the nth element. So, it will be ceil(n/2) for sure. 0 0 replyShare Please log in or register to add a comment.