Mixing
Prove that the quantity is an integer
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I want to prove that $fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $fracn^33-fracn^22+fracn6=frac13-frac12+frac16=0 in mathbbZ$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $frack^33-frack^22+frack6 in mathbbZ$.
Induction step: We want to show that it holds for $n=k+1$.
$$frac(k+1)^33-frac(k+1)^22+frack+16=frack^33+frack^22+frack6$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
elementary-number-theory induction
edited 3 hours ago
amWhy
191k27223434
asked 3 hours ago
Evinda
561412
add a comment |Â
up vote
1
down vote
favorite
I want to prove that $fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $fracn^33-fracn^22+fracn6=frac13-frac12+frac16=0 in mathbbZ$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $frack^33-frack^22+frack6 in mathbbZ$.
Induction step: We want to show that it holds for $n=k+1$.
$$frac(k+1)^33-frac(k+1)^22+frack+16=frack^33+frack^22+frack6$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
elementary-number-theory induction
edited 3 hours ago
amWhy
191k27223434
asked 3 hours ago
Evinda
561412
Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I want to prove that $fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $fracn^33-fracn^22+fracn6=frac13-frac12+frac16=0 in mathbbZ$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $frack^33-frack^22+frack6 in mathbbZ$.
Induction step: We want to show that it holds for $n=k+1$.
$$frac(k+1)^33-frac(k+1)^22+frack+16=frack^33+frack^22+frack6$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
elementary-number-theory induction
edited 3 hours ago
amWhy
191k27223434
asked 3 hours ago
Evinda
561412
I want to prove that $fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $fracn^33-fracn^22+fracn6=frac13-frac12+frac16=0 in mathbbZ$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $frack^33-frack^22+frack6 in mathbbZ$.
Induction step: We want to show that it holds for $n=k+1$.
$$frac(k+1)^33-frac(k+1)^22+frack+16=frack^33+frack^22+frack6$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
elementary-number-theory induction
elementary-number-theory induction
edited 3 hours ago
amWhy
191k27223434
asked 3 hours ago
Evinda
561412
edited 3 hours ago
amWhy
191k27223434
asked 3 hours ago
Evinda
561412
edited 3 hours ago
amWhy
191k27223434
edited 3 hours ago
amWhy
191k27223434
edited 3 hours ago
amWhy
191k27223434
191k27223434
asked 3 hours ago
Evinda
561412
asked 3 hours ago
Evinda
561412
asked 3 hours ago
Evinda
561412
561412
Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago
add a comment |Â
Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago
Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
2
down vote
accepted
If you want to use induction you want to show that
$$frac(k+1)^33-frac(k+1)^22+frack+16$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
answered 3 hours ago
Ross Millikan
283k23191359
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
Yes, we can. If $f(n)=fracn^33-fracn^22+fracn6$, then $f(n+1)-f(n)=n^2$.
answered 3 hours ago
metamorphy
1,609111
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
$$fracn^33-fracn^22+fracn6=fracn(n-1)(2n-1)6$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer.
answered 3 hours ago
sirous
1,108513
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
To prove $$fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
If you put everything over a common denominator you get $$f(n)=frac 2n^3-3n^2+n6=frac n(n-1)(2n-1)6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=frac n(n-1)(2n-4+3)6=frac n(n-1)(n-2)3+frac n(n-1)2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
answered 3 hours ago
Mark Bennet
78.4k777177
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If you want to use induction you want to show that
$$frac(k+1)^33-frac(k+1)^22+frack+16$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
answered 3 hours ago
Ross Millikan
283k23191359
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
If you want to use induction you want to show that
$$frac(k+1)^33-frac(k+1)^22+frack+16$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
answered 3 hours ago
Ross Millikan
283k23191359
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If you want to use induction you want to show that
$$frac(k+1)^33-frac(k+1)^22+frack+16$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
answered 3 hours ago
Ross Millikan
283k23191359
If you want to use induction you want to show that
$$frac(k+1)^33-frac(k+1)^22+frack+16$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.
Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
answered 3 hours ago
Ross Millikan
283k23191359
283k23191359
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
add a comment |Â
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Why is one of the terms a multiple of $3$ ?
– Evinda
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Just try the three possibilities for $k bmod 3$. If it is $0$, then $k$ is a multiple of $3$. If $1$, then $k-1$ is. And if $2$, then $2k-1$ is.
– Ross Millikan
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
Ah, I see... And so we have that $2 mid (2k-1)(k-1)k$ and $3 mid (2k-1)(k-1)k$ and thus $2 cdot 3=6 mid (2k-1)(k-1)k$, right?
– Evinda
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
That is correct.
– Ross Millikan
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
Great, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
Yes, we can. If $f(n)=fracn^33-fracn^22+fracn6$, then $f(n+1)-f(n)=n^2$.
answered 3 hours ago
metamorphy
1,609111
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
Yes, we can. If $f(n)=fracn^33-fracn^22+fracn6$, then $f(n+1)-f(n)=n^2$.
answered 3 hours ago
metamorphy
1,609111
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Yes, we can. If $f(n)=fracn^33-fracn^22+fracn6$, then $f(n+1)-f(n)=n^2$.
answered 3 hours ago
metamorphy
1,609111
Yes, we can. If $f(n)=fracn^33-fracn^22+fracn6$, then $f(n+1)-f(n)=n^2$.
answered 3 hours ago
metamorphy
1,609111
answered 3 hours ago
metamorphy
1,609111
answered 3 hours ago
metamorphy
1,609111
answered 3 hours ago
metamorphy
1,609111
1,609111
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
add a comment |Â
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
And since $f(n)$ is an integer and also $n^2$, we deduce that their sum, $f(n+1)=f(n)+n^2$, is also an integer, right? @metamorphy
– Evinda
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Exactly. There are other approaches noted above - I've just followed your way.
– metamorphy
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
Nice, thank you :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
$$fracn^33-fracn^22+fracn6=fracn(n-1)(2n-1)6$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer.
answered 3 hours ago
sirous
1,108513
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
$$fracn^33-fracn^22+fracn6=fracn(n-1)(2n-1)6$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer.
answered 3 hours ago
sirous
1,108513
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$$fracn^33-fracn^22+fracn6=fracn(n-1)(2n-1)6$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer.
answered 3 hours ago
sirous
1,108513
$$fracn^33-fracn^22+fracn6=fracn(n-1)(2n-1)6$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer.
answered 3 hours ago
sirous
1,108513
answered 3 hours ago
sirous
1,108513
answered 3 hours ago
sirous
1,108513
answered 3 hours ago
sirous
1,108513
1,108513
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
I see... Thanks :)
– Evinda
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
To prove $$fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
To prove $$fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
To prove $$fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
To prove $$fracn^33-fracn^22+fracn6 in mathbbZ, forall n geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
answered 3 hours ago
Mohammad Riazi-Kermani
33.9k41855
33.9k41855
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
I haven't understood why the product is divisible by 3. Could you explain it to me?
– Evinda
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
We have $n(n-1)(2n-1)$. For$ n=3k$ , $n$ is a multiple of $3$ so the product is a multiple of $3$. For $n=3k+1$ , $n-1$ is a multiple of $3$ and for $n=3k+2$, $2n-1 =6k+3$ which is a multiple of $3$.
– Mohammad Riazi-Kermani
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
I see... Thanks :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
If you put everything over a common denominator you get $$f(n)=frac 2n^3-3n^2+n6=frac n(n-1)(2n-1)6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=frac n(n-1)(2n-4+3)6=frac n(n-1)(n-2)3+frac n(n-1)2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
answered 3 hours ago
Mark Bennet
78.4k777177
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
If you put everything over a common denominator you get $$f(n)=frac 2n^3-3n^2+n6=frac n(n-1)(2n-1)6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=frac n(n-1)(2n-4+3)6=frac n(n-1)(n-2)3+frac n(n-1)2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
answered 3 hours ago
Mark Bennet
78.4k777177
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you put everything over a common denominator you get $$f(n)=frac 2n^3-3n^2+n6=frac n(n-1)(2n-1)6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=frac n(n-1)(2n-4+3)6=frac n(n-1)(n-2)3+frac n(n-1)2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
answered 3 hours ago
Mark Bennet
78.4k777177
If you put everything over a common denominator you get $$f(n)=frac 2n^3-3n^2+n6=frac n(n-1)(2n-1)6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=frac n(n-1)(2n-4+3)6=frac n(n-1)(n-2)3+frac n(n-1)2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
answered 3 hours ago
Mark Bennet
78.4k777177
answered 3 hours ago
Mark Bennet
78.4k777177
answered 3 hours ago
Mark Bennet
78.4k777177
answered 3 hours ago
Mark Bennet
78.4k777177
78.4k777177
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
add a comment |Â
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
I like this approach. Thanks a lot :)
– Evinda
3 hours ago
add a comment |Â
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Why would you think those expressions would be equal? The induction step would be to show that the integrality of the right hand implied that of the left hand.
– lulu
3 hours ago
(I've put an answer as it is almost done already.)
– metamorphy
3 hours ago
Side note: I think it is easier to first argue that it's enough to check $nin 0,1,2,3,4,5$ and then check those cases. Or, alternatively, note that your expression is $frac 16 times n(n-1)(2n-1)$ and show, by cases, that at least one of the terms in $n$ is even and at least one is divisible by $3$.
– lulu
3 hours ago