BdMO 2020 Preliminary Higher Secondary P4

Latest News, Announcements, and Forum Rules
mdvirus
Posts:1
Joined:Sat Feb 20, 2021 7:13 pm
BdMO 2020 Preliminary Higher Secondary P4

Unread post by mdvirus » Thu Apr 08, 2021 5:50 pm

If $n$ is even, then $T(n) = T(n - 1) + 1$ and if $n$ is odd then $T(n) = T(n - 2) + 2$. If $T(1) = 7$ what is $T(2020)$?

Asif Hossain
Posts:193
Joined:Sat Jan 02, 2021 9:28 pm

Re: BdMO 2020 Preliminary Higher Secondary P4

Unread post by Asif Hossain » Sun Apr 25, 2021 10:35 am

You should have posted the problem in BdMO problem section but anyways
If $n+1$ is odd then $T(n+1)=T(n-1)+2 \Rightarrow T(n+1)=T(n)-1+2 \Rightarrow T(n+1)=T(n)+1$
If $n+1$ is even then $T(n+1)=T(n)+1$ so $T(n)$ is in arithmetic progression so $T(2020)=T(1)+(2020-1)=2026$
Hmm..Hammer...Treat everything as nail

niamul21
Posts:1
Joined:Thu Jun 17, 2021 8:19 am

Re: BdMO 2020 Preliminary Higher Secondary P4

Unread post by niamul21 » Thu Jun 17, 2021 10:03 am

mdvirus wrote:
Thu Apr 08, 2021 5:50 pm
If $n$ is even, then $T(n) = T(n - 1) + 1$ and if $n$ is odd then $T(n) = T(n - 2) + 2$. If $T(1) = 7$ what is $T(2020)$?
You did it right. Thanks

Post Reply