Proof by mathematical induction with the problem $40(2n)! ≥ 30^n$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Proof by Induction (concerning $3^nge1+2n$)Mathematical Induction with InequalitiesProve by using Mathematical induction (sum of the first $n$ odd numbers is $n^2$)Stuck in Induction Inequality: $2^n>3n^2$prove inequality by induction — Discrete mathProve by induction $n! > n^2$Use mathematical induction to prove the following $n! < n^n$Mathematical induction by inequalityStuck On A Proof By InductionProve by induction, for all positive integers $n$, that $n!ge2^n-1$
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Proof by mathematical induction with the problem $40(2n)! ≥ 30^n$
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Proof by Induction (concerning $3^nge1+2n$)Mathematical Induction with InequalitiesProve by using Mathematical induction (sum of the first $n$ odd numbers is $n^2$)Stuck in Induction Inequality: $2^n>3n^2$prove inequality by induction — Discrete mathProve by induction $n! > n^2$Use mathematical induction to prove the following $n! < n^n$Mathematical induction by inequalityStuck On A Proof By InductionProve by induction, for all positive integers $n$, that $n!ge2^n-1$
$begingroup$
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
edited 3 mins ago
YuiTo Cheng
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
$endgroup$
add a comment |
$begingroup$
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
edited 3 mins ago
YuiTo Cheng
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited 3 mins ago
YuiTo Cheng
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
$endgroup$
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
edited 3 mins ago
YuiTo Cheng
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
edited 3 mins ago
YuiTo Cheng
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
edited 3 mins ago
YuiTo Cheng
2,58641037
edited 3 mins ago
YuiTo Cheng
2,58641037
edited 3 mins ago
YuiTo Cheng
2,58641037
2,58641037
asked 2 hours ago
Nick SabiaNick Sabia
566
asked 2 hours ago
Nick SabiaNick Sabia
566
asked 2 hours ago
Nick SabiaNick Sabia
566
566
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
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.
answered 1 hour ago
user1952500user1952500
1,133812
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
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.
answered 1 hour ago
user1952500user1952500
1,133812
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
user1952500user1952500
1,133812
$endgroup$
add a comment |
$begingroup$
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.
answered 1 hour ago
user1952500user1952500
1,133812
$endgroup$
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.
answered 1 hour ago
user1952500user1952500
1,133812
answered 1 hour ago
user1952500user1952500
1,133812
answered 1 hour ago
user1952500user1952500
1,133812
answered 1 hour ago
user1952500user1952500
1,133812
1,133812
add a comment |
add a comment |
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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