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Posers and Puzzles

Posers and Puzzles

  1. 25 Jan '05 18:29 / 1 edit
    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?
  2. Standard member DoctorScribbles
    BWA Soldier
    25 Jan '05 21:29
    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.
  3. 26 Jan '05 05:52
    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?
  4. Standard member DoctorScribbles
    BWA Soldier
    26 Jan '05 07:04
    Originally posted by sarathian
    I think I too have the proof. Can you give detailed steps of YOUR proof?
    Induction does the trick.
  5. 26 Jan '05 08:56
    Yep, Induction is the easiest way.
  6. Standard member TheMaster37
    Kupikupopo!
    26 Jan '05 10:24
    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.
  7. 26 Jan '05 12:48 / 1 edit
    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.
  8. Standard member TheMaster37
    Kupikupopo!
    26 Jan '05 13:53
    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".
  9. 26 Jan '05 19:58
    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