text: # Prove by induction
  
  1. Sep 15, 2009
  
  ### rbnphlp
  
  3n>n3 where n >3
  
  I know I have to use proof by induction to solve this.
  assume for f(4)is true
  for f(n+1)=3n+1>3n3
  
  However after that I dont have a clue of how to getting it into (n+1)3
  Any hints will be greatly apperciated
  
  2. Sep 15, 2009
  
  Let
  
  $$A_n$$
  
  be the statement
  
  $$3^n > n^3$$
  
  You want to show $$A_n$$ is true for $$n \ge 4$$.
  
  It shouldn't be hard to show $$A_4$$ is true. Assume it is true for $$k \ge 4$$.
  
  Now
  
  $$3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3$$
  
  Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of $$(k+1)^3$$.
  
  (Hint for a start: You know by hypothesis $$k \ge 4 > 3$$, so
  $$k^3 = k \cdot k^2 > 3k^2$$
  which is the second term in the expansion of $$(k+1)^3$$)
  
  Last edited by a moderator: Sep 15, 2009
  3. Sep 16, 2009
  
  ### rbnphlp
  
  This is th best I could get at:
  
  3n>n3
  For n+1=
  3n+1>3n
  
  given n>3 , I found the following:
  
  • n3>3n2
  • n2>3n
  • n-2>1
  with the best intentions I did the following:
  
  Added all them up and got
  n3>3n2+3n+1-n2-n+2--(1)
  3n+1-2n3>n3--(2)
  
  from 1 and 2
  3n+1-2n3>3n2+3n+1-n2-n+2
  
  3n+1+n2+n-2-n3>(n+1)3
  
  But after that im stuck it would be great some one could help. thnx
  
  4. Sep 16, 2009
  
  ### rbnphlp
  
  anyone?
  
  5. Sep 16, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  Statdad showed that you could get to $k^3+ k^3+ k^3$ and you want to compare that to $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Can you show that $k^3> 3k^2$ when k> 3? Can you show that $k^3> 3k+ 1$ when k> 3?
  
  6. Sep 17, 2009
  
  ### rbnphlp
  
  nope I cant .
explanation: The document is a discussion thread from a forum where a user is seeking help with a proof by induction problem. The problem is to prove that \(3^n > n^3\) for \(n > 3\). The thread contains a question from the user and multiple responses providing guidance and hints to solve the problem.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Prove by induction
  
  1. Sep 15, 2009
  
  ### rbnphlp
  
  3n>n3 where n >3
  
  I know I have to use proof by induction to solve this.
  assume for f(4)is true
  for f(n+1)=3n+1>3n3
  
  However after that I dont have a clue of how to getting it into (n+1)3
  Any hints will be greatly apperciated
  
  2. Sep 15, 2009
  
  Let
  
  $$A_n$$
  
  be the statement
  
  $$3^n > n^3$$
  
  You want to show $$A_n$$ is true for $$n \ge 4$$.
  
  It shouldn't be hard to show $$A_4$$ is true. Assume it is true for $$k \ge 4$$.
  
  Now
  
  $$3^{3+1} = 3 \cdot 3 ^k > 3 \cdot k^3 = k^3 + k^3 + k^3$$
  
  Now play with the terms on the right, making the new expressions ever smaller, to build up to the expansion of $$(k+1)^3$$.
  
  (Hint for a start: You know by hypothesis $$k \ge 4 > 3$$, so
  $$k^3 = k \cdot k^2 > 3k^2$$
  which is the second term in the expansion of $$(k+1)^3$$)
  
  Last edited by a moderator: Sep 15, 2009
  3. Sep 16, 2009
  
  ### rbnphlp
  
  This is th best I could get at:
  
  3n>n3
  For n+1=
  3n+1>3n
  
  given n>3 , I found the following:
  
  • n3>3n2
  • n2>3n
  • n-2>1
  with the best intentions I did the following:
  
  Added all them up and got
  n3>3n2+3n+1-n2-n+2--(1)
  3n+1-2n3>n3--(2)
  
  from 1 and 2
  3n+1-2n3>3n2+3n+1-n2-n+2
  
  3n+1+n2+n-2-n3>(n+1)3
  
  But after that im stuck it would be great some one could help. thnx
  
  4. Sep 16, 2009
  
  ### rbnphlp
  
  anyone?
  
  5. Sep 16, 2009
  
  ### HallsofIvy
  
  Staff Emeritus
  Statdad showed that you could get to $k^3+ k^3+ k^3$ and you want to compare that to $(k+1)^3= k^3+ 3k^2+ 3k+ 1$. Can you show that $k^3> 3k^2$ when k> 3? Can you show that $k^3> 3k+ 1$ when k> 3?
  
  6. Sep 17, 2009
  
  ### rbnphlp
  
  nope I cant .
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a discussion thread from a forum where a user is seeking help with a proof by induction problem. The problem is to prove that \(3^n > n^3\) for \(n > 3\). The thread contains a question from the user and multiple responses providing guidance and hints to solve the problem.
  
  Label: question-and-answer