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












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$begingroup$


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







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  • $begingroup$
    For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
    $endgroup$
    – 1123581321
    2 mins ago


















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











$endgroup$












  • $begingroup$
    For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
    $endgroup$
    – 1123581321
    2 mins ago
















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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edited 3 mins ago









YuiTo Cheng

2,58641037




2,58641037










asked 2 hours ago









Nick SabiaNick Sabia

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  • $begingroup$
    For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
    $endgroup$
    – 1123581321
    2 mins ago




















  • $begingroup$
    For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
    $endgroup$
    – 1123581321
    2 mins ago


















$begingroup$
For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
$endgroup$
– 1123581321
2 mins ago






$begingroup$
For induction proofs involving an inequality, the restriction on the values of $n$ is almost always going to form part of the proof (see the answer below).
$endgroup$
– 1123581321
2 mins ago












1 Answer
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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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      4












      $begingroup$

      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









      $endgroup$
















        4












        4








        4





        $begingroup$

        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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