Prove using induction that $n^6 18$











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Prove using induction (or using any other elementary precalculus techniques) that $$n^6 < 3^n, forall n geq 19.$$




I have no idea how to do this. Writing the induction step, I get that I need to prove that $$3n^6 > (n+1)^6,$$ and I don't know how to do so.



I want a proof that doesn't use calculus techniques.






inequality induction natural-numbers







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    what do you mean by 'calculus techniques'? is induction allowed?
    – Viktor Glombik
    yesterday















up vote
-2
down vote

favorite













Prove using induction (or using any other elementary precalculus techniques) that $$n^6 < 3^n, forall n geq 19.$$




I have no idea how to do this. Writing the induction step, I get that I need to prove that $$3n^6 > (n+1)^6,$$ and I don't know how to do so.



I want a proof that doesn't use calculus techniques.






inequality induction natural-numbers







share|cite|improve this question


















  • 1




    what do you mean by 'calculus techniques'? is induction allowed?
    – Viktor Glombik
    yesterday













up vote
-2
down vote

favorite









up vote
-2
down vote

favorite












Prove using induction (or using any other elementary precalculus techniques) that $$n^6 < 3^n, forall n geq 19.$$




I have no idea how to do this. Writing the induction step, I get that I need to prove that $$3n^6 > (n+1)^6,$$ and I don't know how to do so.



I want a proof that doesn't use calculus techniques.






inequality induction natural-numbers







share|cite|improve this question














Prove using induction (or using any other elementary precalculus techniques) that $$n^6 < 3^n, forall n geq 19.$$




I have no idea how to do this. Writing the induction step, I get that I need to prove that $$3n^6 > (n+1)^6,$$ and I don't know how to do so.



I want a proof that doesn't use calculus techniques.






inequality induction natural-numbers




inequality induction natural-numbers






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked yesterday









S.T.

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  • 1




    what do you mean by 'calculus techniques'? is induction allowed?
    – Viktor Glombik
    yesterday














  • 1




    what do you mean by 'calculus techniques'? is induction allowed?
    – Viktor Glombik
    yesterday








1




1




what do you mean by 'calculus techniques'? is induction allowed?
– Viktor Glombik
yesterday




what do you mean by 'calculus techniques'? is induction allowed?
– Viktor Glombik
yesterday










2 Answers
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Use that



$$3^{n+1}=3cdot 3^nstackrel{Ind. Hyp.}>3cdot n^6 stackrel{?}>(n+1)^6$$



and



$$3cdot n^6 >(n+1)^6 iff frac{n+1}{n}<sqrt[6] 3 iff n>frac1{sqrt[6] 3-1}approx 4.98$$






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    If $nge 19$,



    $$(n+1)^6=n^6left(1+frac1nright)^6le3^ncdot left(1+frac1{19}right)^6<3^ncdot1.1^6$$






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
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      up vote
      4
      down vote













      Use that



      $$3^{n+1}=3cdot 3^nstackrel{Ind. Hyp.}>3cdot n^6 stackrel{?}>(n+1)^6$$



      and



      $$3cdot n^6 >(n+1)^6 iff frac{n+1}{n}<sqrt[6] 3 iff n>frac1{sqrt[6] 3-1}approx 4.98$$






      share|cite|improve this answer



























        up vote
        4
        down vote













        Use that



        $$3^{n+1}=3cdot 3^nstackrel{Ind. Hyp.}>3cdot n^6 stackrel{?}>(n+1)^6$$



        and



        $$3cdot n^6 >(n+1)^6 iff frac{n+1}{n}<sqrt[6] 3 iff n>frac1{sqrt[6] 3-1}approx 4.98$$






        share|cite|improve this answer

























          up vote
          4
          down vote










          up vote
          4
          down vote









          Use that



          $$3^{n+1}=3cdot 3^nstackrel{Ind. Hyp.}>3cdot n^6 stackrel{?}>(n+1)^6$$



          and



          $$3cdot n^6 >(n+1)^6 iff frac{n+1}{n}<sqrt[6] 3 iff n>frac1{sqrt[6] 3-1}approx 4.98$$






          share|cite|improve this answer














          Use that



          $$3^{n+1}=3cdot 3^nstackrel{Ind. Hyp.}>3cdot n^6 stackrel{?}>(n+1)^6$$



          and



          $$3cdot n^6 >(n+1)^6 iff frac{n+1}{n}<sqrt[6] 3 iff n>frac1{sqrt[6] 3-1}approx 4.98$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited yesterday

























          answered yesterday









          gimusi

          85.6k74294




          85.6k74294






















              up vote
              1
              down vote













              If $nge 19$,



              $$(n+1)^6=n^6left(1+frac1nright)^6le3^ncdot left(1+frac1{19}right)^6<3^ncdot1.1^6$$






              share|cite|improve this answer

























                up vote
                1
                down vote













                If $nge 19$,



                $$(n+1)^6=n^6left(1+frac1nright)^6le3^ncdot left(1+frac1{19}right)^6<3^ncdot1.1^6$$






                share|cite|improve this answer























                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  If $nge 19$,



                  $$(n+1)^6=n^6left(1+frac1nright)^6le3^ncdot left(1+frac1{19}right)^6<3^ncdot1.1^6$$






                  share|cite|improve this answer












                  If $nge 19$,



                  $$(n+1)^6=n^6left(1+frac1nright)^6le3^ncdot left(1+frac1{19}right)^6<3^ncdot1.1^6$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered yesterday









                  ajotatxe

                  52.1k23688




                  52.1k23688






























                       

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