Proof by mathematical induction with the problem 40(2n)! ≥ 30^n












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I want to start by saying that I have for less trouble handling a non-inequality induction problem. I really don't understand the steps to take to get to the desired end product with these inequality induction proofs. That being said, I feel like I just wrote a mess on my paper that leads me nowhere. Here's my proof so far for the mathematical induction of $40(2n)!≥30^{n}$, where n ≥ 1



Let P(n) be the statement $40(2n)!≥30^{n}$ where n ≥ 1.



Basis Step: (n = 1) $LHS = 40 * 2! = 80$ and $RHS = 30^{1} = 30$



$80 ≥ 30$



Inductive Step: Assume $P(k)$ is true for $k = 1$. Our goal is to show $P(k+1)$ is true by showing $40(2(k+1))!≥30^{k+1}$ for $k ≥ 1$.



(Beyond this step I have no clue how to alter the LHS factorial or the RHS exponent in such a way to benefit me. Below is my work so far)



LHS: (I tried to multiply $(2k+2)(2k+1)$ to both sides, but then I didn't see how that would help)$40(2k)!≥30^{k} = (2k)!*40≥30^{k} = (2k+2)(2k+1)(2k)!*40≥30^{k}*(2k+2)(2k+1)$



RHS: $40(2(k+1))!≥30^{k+1}=(2k+2)!*40≥30^{k+1}=(2k+2)(2k+1)(2k)!*40≥30^{k+1}$ $=(2k+2)(2k+1)(2k)!*40≥30*30^{k}$



(I then assumed the inductive hypothesis and placed $40(2k)!≥30^{k}$ in the middle to get...)



$(2k+2)(2k+1)(2k)!*40≥40(2k)!≥30^{k}≥30*30^{k}$



At this point, I've got nothing. I know that $30^{k}≥30*30^{k}$ makes no sense, but I don't know where to move $30*30^{k}$ since I can't assume that $40(2k)!≥30*30^{k}$. I don't have any clue how I can manipulate either side to help me.






inequality induction







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    1












    $begingroup$


    I want to start by saying that I have for less trouble handling a non-inequality induction problem. I really don't understand the steps to take to get to the desired end product with these inequality induction proofs. That being said, I feel like I just wrote a mess on my paper that leads me nowhere. Here's my proof so far for the mathematical induction of $40(2n)!≥30^{n}$, where n ≥ 1



    Let P(n) be the statement $40(2n)!≥30^{n}$ where n ≥ 1.



    Basis Step: (n = 1) $LHS = 40 * 2! = 80$ and $RHS = 30^{1} = 30$



    $80 ≥ 30$



    Inductive Step: Assume $P(k)$ is true for $k = 1$. Our goal is to show $P(k+1)$ is true by showing $40(2(k+1))!≥30^{k+1}$ for $k ≥ 1$.



    (Beyond this step I have no clue how to alter the LHS factorial or the RHS exponent in such a way to benefit me. Below is my work so far)



    LHS: (I tried to multiply $(2k+2)(2k+1)$ to both sides, but then I didn't see how that would help)$40(2k)!≥30^{k} = (2k)!*40≥30^{k} = (2k+2)(2k+1)(2k)!*40≥30^{k}*(2k+2)(2k+1)$



    RHS: $40(2(k+1))!≥30^{k+1}=(2k+2)!*40≥30^{k+1}=(2k+2)(2k+1)(2k)!*40≥30^{k+1}$ $=(2k+2)(2k+1)(2k)!*40≥30*30^{k}$



    (I then assumed the inductive hypothesis and placed $40(2k)!≥30^{k}$ in the middle to get...)



    $(2k+2)(2k+1)(2k)!*40≥40(2k)!≥30^{k}≥30*30^{k}$



    At this point, I've got nothing. I know that $30^{k}≥30*30^{k}$ makes no sense, but I don't know where to move $30*30^{k}$ since I can't assume that $40(2k)!≥30*30^{k}$. I don't have any clue how I can manipulate either side to help me.






    inequality induction







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I want to start by saying that I have for less trouble handling a non-inequality induction problem. I really don't understand the steps to take to get to the desired end product with these inequality induction proofs. That being said, I feel like I just wrote a mess on my paper that leads me nowhere. Here's my proof so far for the mathematical induction of $40(2n)!≥30^{n}$, where n ≥ 1



      Let P(n) be the statement $40(2n)!≥30^{n}$ where n ≥ 1.



      Basis Step: (n = 1) $LHS = 40 * 2! = 80$ and $RHS = 30^{1} = 30$



      $80 ≥ 30$



      Inductive Step: Assume $P(k)$ is true for $k = 1$. Our goal is to show $P(k+1)$ is true by showing $40(2(k+1))!≥30^{k+1}$ for $k ≥ 1$.



      (Beyond this step I have no clue how to alter the LHS factorial or the RHS exponent in such a way to benefit me. Below is my work so far)



      LHS: (I tried to multiply $(2k+2)(2k+1)$ to both sides, but then I didn't see how that would help)$40(2k)!≥30^{k} = (2k)!*40≥30^{k} = (2k+2)(2k+1)(2k)!*40≥30^{k}*(2k+2)(2k+1)$



      RHS: $40(2(k+1))!≥30^{k+1}=(2k+2)!*40≥30^{k+1}=(2k+2)(2k+1)(2k)!*40≥30^{k+1}$ $=(2k+2)(2k+1)(2k)!*40≥30*30^{k}$



      (I then assumed the inductive hypothesis and placed $40(2k)!≥30^{k}$ in the middle to get...)



      $(2k+2)(2k+1)(2k)!*40≥40(2k)!≥30^{k}≥30*30^{k}$



      At this point, I've got nothing. I know that $30^{k}≥30*30^{k}$ makes no sense, but I don't know where to move $30*30^{k}$ since I can't assume that $40(2k)!≥30*30^{k}$. I don't have any clue how I can manipulate either side to help me.






      inequality induction







      share|cite|improve this question









      $endgroup$




      I want to start by saying that I have for less trouble handling a non-inequality induction problem. I really don't understand the steps to take to get to the desired end product with these inequality induction proofs. That being said, I feel like I just wrote a mess on my paper that leads me nowhere. Here's my proof so far for the mathematical induction of $40(2n)!≥30^{n}$, where n ≥ 1



      Let P(n) be the statement $40(2n)!≥30^{n}$ where n ≥ 1.



      Basis Step: (n = 1) $LHS = 40 * 2! = 80$ and $RHS = 30^{1} = 30$



      $80 ≥ 30$



      Inductive Step: Assume $P(k)$ is true for $k = 1$. Our goal is to show $P(k+1)$ is true by showing $40(2(k+1))!≥30^{k+1}$ for $k ≥ 1$.



      (Beyond this step I have no clue how to alter the LHS factorial or the RHS exponent in such a way to benefit me. Below is my work so far)



      LHS: (I tried to multiply $(2k+2)(2k+1)$ to both sides, but then I didn't see how that would help)$40(2k)!≥30^{k} = (2k)!*40≥30^{k} = (2k+2)(2k+1)(2k)!*40≥30^{k}*(2k+2)(2k+1)$



      RHS: $40(2(k+1))!≥30^{k+1}=(2k+2)!*40≥30^{k+1}=(2k+2)(2k+1)(2k)!*40≥30^{k+1}$ $=(2k+2)(2k+1)(2k)!*40≥30*30^{k}$



      (I then assumed the inductive hypothesis and placed $40(2k)!≥30^{k}$ in the middle to get...)



      $(2k+2)(2k+1)(2k)!*40≥40(2k)!≥30^{k}≥30*30^{k}$



      At this point, I've got nothing. I know that $30^{k}≥30*30^{k}$ makes no sense, but I don't know where to move $30*30^{k}$ since I can't assume that $40(2k)!≥30*30^{k}$. I don't have any clue how I can manipulate either side to help me.






      inequality induction




      inequality induction






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      asked 1 hour ago









      Nick SabiaNick Sabia

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          For $k ge 2$, $(2k+1)(2k+2) ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,



          $$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$
          $$ge 30^k (2k+1)(2k+2)quadquadquad (since 40(2k)! >= 30^k by P(k))$$
          $$ ge 30^k cdot 30 forall k ge 2$$
          $$ ge 30^{k+1}$$



          So you need to verify the proposition for $k=2$ and proceed with the induction.






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            $begingroup$

            For $k ge 2$, $(2k+1)(2k+2) ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,



            $$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$
            $$ge 30^k (2k+1)(2k+2)quadquadquad (since 40(2k)! >= 30^k by P(k))$$
            $$ ge 30^k cdot 30 forall k ge 2$$
            $$ ge 30^{k+1}$$



            So you need to verify the proposition for $k=2$ and proceed with the induction.






            share|cite|improve this answer









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              $begingroup$

              For $k ge 2$, $(2k+1)(2k+2) ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,



              $$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$
              $$ge 30^k (2k+1)(2k+2)quadquadquad (since 40(2k)! >= 30^k by P(k))$$
              $$ ge 30^k cdot 30 forall k ge 2$$
              $$ ge 30^{k+1}$$



              So you need to verify the proposition for $k=2$ and proceed with the induction.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                For $k ge 2$, $(2k+1)(2k+2) ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,



                $$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$
                $$ge 30^k (2k+1)(2k+2)quadquadquad (since 40(2k)! >= 30^k by P(k))$$
                $$ ge 30^k cdot 30 forall k ge 2$$
                $$ ge 30^{k+1}$$



                So you need to verify the proposition for $k=2$ and proceed with the induction.






                share|cite|improve this answer









                $endgroup$



                For $k ge 2$, $(2k+1)(2k+2) ge 30$. Hence if $P(k)$ is true for $k=1$ and $k=2$,



                $$P(k+1) = 40(2k+2)! = 40(2k)! (2k+1)(2k+2)$$
                $$ge 30^k (2k+1)(2k+2)quadquadquad (since 40(2k)! >= 30^k by P(k))$$
                $$ ge 30^k cdot 30 forall k ge 2$$
                $$ ge 30^{k+1}$$



                So you need to verify the proposition for $k=2$ and proceed with the induction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                user1952500user1952500

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                1,123812






























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