Mystery sequence generator - who can explain ?

Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.


Can someone prove as to why is this so? And is this true for all n?

Originally posted by ranjan sinha
Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.


Can someone prove as to why is this so? And is this true for all n?

I can prove that it is true for all n.

Originally posted by DoctorScribbles
I can prove that it is true for all n.

I think I too have the proof. Can you give detailed steps of YOUR proof?

Originally posted by sarathian
I think I too have the proof. Can you give detailed steps of YOUR proof?

Induction does the trick.

Yep, Induction is the easiest way.

Originally posted by ranjan sinha
Consider the sequence of integers s(1), s(2), s(3), .... s(n)....... defined by s(1)= 1; s(n+1) = 3*s(n) +1:etc..
Now it so turns out that 2*s(n) +1 = 3^n.
I have veriffied this to be true for n= 1,2,3,4, etc.


Can someone prove as to why is this so? And is this true for all n?

Say above holds for a number N, define M = N+1

Then
2*s(M) + 1 =
2*(3*s(N) + 1) + 1 =
2*(s(N) + 2*s(N) + 1) + 1 =
2*(s(N) + 3^N) + 1 =
2*s(N) + 2*3^N + 1 =
3^N + 2*3^N =
3*3^N =
3^M

So if above holds for N, it also holds for M. Above holds for 1.

Originally posted by TheMaster37
Say above holds for a number N, define M = N+1

Then
2*s(M) + 1 =
2*(3*s(N) + 1) + 1 =
2*(s(N) + 2*s(N) + 1) + 1 =
2*(s(N) + 3^N) + 1 =
2*s(N) + 2*3^N + 1 =
3^N + 2*3^N =
3*3^N =
3^M

So if above holds for N, it also holds for M. Above holds for 1.

Yep...That's it.
🙄🙄😲😲

I was jsut feeling like writing it out to stay in practise. I generally find it very lazy to simply say "I have the proof".

Originally posted by TheMaster37
I was jsut feeling like writing it out to stay in practise. I generally find it very lazy to simply say "I have the proof".

You're definitely out of practise 😛

Much shorter (only one substitution) is:

1 + 2*s(M) =
1 + (6*s(N) + 2) =
3 * (2*s(N) + 1) =
3 * 3^N =
3^M