- 25 Jan '05 18:29 / 1 editConsider 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? - 25 Jan '05 21:29

I can prove that it is 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? - 26 Jan '05 10:24

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

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. - 26 Jan '05 12:48 / 1 edit

Yep...That's it.*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. - 26 Jan '05 19:58

You're definitely out of practise*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".**

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