BdMO 2020 Preliminary Higher Secondary P4
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 $T(2020)$?
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Re: BdMO 2020 Preliminary Higher Secondary P4
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 $T(2020)=T(1)+(2020-1)=2026$
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