Mixing
Using Mathematical Induction To Prove
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I am having trouble proving the following using Mathematical Induction.
1,3,5...(2n-1)∕2,4,6...(2n) ≥ 1∕2n
I cannot seem to understand using Induction to prove a fractional expression such as the one above.
I assume the proof for the base case of n is required, like most induction questions.
However I do seem to be lost with this one.
All help is appreciated.
induction
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
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I am having trouble proving the following using Mathematical Induction.
1,3,5...(2n-1)∕2,4,6...(2n) ≥ 1∕2n
I cannot seem to understand using Induction to prove a fractional expression such as the one above.
I assume the proof for the base case of n is required, like most induction questions.
However I do seem to be lost with this one.
All help is appreciated.
induction
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am having trouble proving the following using Mathematical Induction.
1,3,5...(2n-1)∕2,4,6...(2n) ≥ 1∕2n
I cannot seem to understand using Induction to prove a fractional expression such as the one above.
I assume the proof for the base case of n is required, like most induction questions.
However I do seem to be lost with this one.
All help is appreciated.
induction
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
I am having trouble proving the following using Mathematical Induction.
1,3,5...(2n-1)∕2,4,6...(2n) ≥ 1∕2n
I cannot seem to understand using Induction to prove a fractional expression such as the one above.
I assume the proof for the base case of n is required, like most induction questions.
However I do seem to be lost with this one.
All help is appreciated.
induction
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
asked Jul 27 at 3:38
KeptAwakeByCoffee
102
102
Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43
add a comment |Â
Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43
Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43
Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43
add a comment |Â
1 Answer
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0
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The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $frac12geq frac12$, which is true.
Assume now that the property is true for some $nin mathbbN$. We want to show that it is true for $n+1$. We have
$$prod_k=1^n+1frac2k-12k=frac2n+12(n+1)times prod_k=1^nfrac2k-12k$$
where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get
$$prod_k=1^n+1frac2k-12kgeq frac2n+12(n+1)times frac12n$$
We want this last quantity to be greater or equal $frac12(n+1)$. Thus, we need to show that $frac2n+12ngeq 1$, which actually is obvious. So we conclude
$$prod_k=1^n+1frac2k-12kgeq frac12(n+1)$$
which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.
answered Jul 27 at 3:45
Suzet
2,203427
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $frac12geq frac12$, which is true.
Assume now that the property is true for some $nin mathbbN$. We want to show that it is true for $n+1$. We have
$$prod_k=1^n+1frac2k-12k=frac2n+12(n+1)times prod_k=1^nfrac2k-12k$$
where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get
$$prod_k=1^n+1frac2k-12kgeq frac2n+12(n+1)times frac12n$$
We want this last quantity to be greater or equal $frac12(n+1)$. Thus, we need to show that $frac2n+12ngeq 1$, which actually is obvious. So we conclude
$$prod_k=1^n+1frac2k-12kgeq frac12(n+1)$$
which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.
answered Jul 27 at 3:45
Suzet
2,203427
add a comment |Â
up vote
0
down vote
accepted
The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $frac12geq frac12$, which is true.
Assume now that the property is true for some $nin mathbbN$. We want to show that it is true for $n+1$. We have
$$prod_k=1^n+1frac2k-12k=frac2n+12(n+1)times prod_k=1^nfrac2k-12k$$
where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get
$$prod_k=1^n+1frac2k-12kgeq frac2n+12(n+1)times frac12n$$
We want this last quantity to be greater or equal $frac12(n+1)$. Thus, we need to show that $frac2n+12ngeq 1$, which actually is obvious. So we conclude
$$prod_k=1^n+1frac2k-12kgeq frac12(n+1)$$
which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.
answered Jul 27 at 3:45
Suzet
2,203427
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $frac12geq frac12$, which is true.
Assume now that the property is true for some $nin mathbbN$. We want to show that it is true for $n+1$. We have
$$prod_k=1^n+1frac2k-12k=frac2n+12(n+1)times prod_k=1^nfrac2k-12k$$
where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get
$$prod_k=1^n+1frac2k-12kgeq frac2n+12(n+1)times frac12n$$
We want this last quantity to be greater or equal $frac12(n+1)$. Thus, we need to show that $frac2n+12ngeq 1$, which actually is obvious. So we conclude
$$prod_k=1^n+1frac2k-12kgeq frac12(n+1)$$
which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.
answered Jul 27 at 3:45
Suzet
2,203427
The base case is $n=1$ (as the quantities involved are not defined for $n=0$). The property then states $frac12geq frac12$, which is true.
Assume now that the property is true for some $nin mathbbN$. We want to show that it is true for $n+1$. We have
$$prod_k=1^n+1frac2k-12k=frac2n+12(n+1)times prod_k=1^nfrac2k-12k$$
where we took out the $k=n+1$-term. Now, note that the product on the right is the quantity involved in our property, at rank $n$. We may use the induction hypothesis to get
$$prod_k=1^n+1frac2k-12kgeq frac2n+12(n+1)times frac12n$$
We want this last quantity to be greater or equal $frac12(n+1)$. Thus, we need to show that $frac2n+12ngeq 1$, which actually is obvious. So we conclude
$$prod_k=1^n+1frac2k-12kgeq frac12(n+1)$$
which exactly is the property at rank $n+1$. This concludes the proof, using the principle of induction.
answered Jul 27 at 3:45
Suzet
2,203427
answered Jul 27 at 3:45
Suzet
2,203427
answered Jul 27 at 3:45
Suzet
2,203427
answered Jul 27 at 3:45
Suzet
2,203427
2,203427
add a comment |Â
add a comment |Â
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Hint: the inequality is the same as $1times3times5timescdotstimes(2n-1)ge2times4times6timescdotstimes(2n-2)$, which you may find easier.
– David
Jul 27 at 3:43