A formula for the term of a sequence
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0
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A seemingly simple but a challenging quiz,
find the nth term in 1,6,30,120.....
I am thinking is a quadratic due to the constant difference but n th term is needed not a recursive one - I got confused
numerical-methods
asked Nov 15 at 1:30
Waitara Mburu
262
add a comment |
up vote
0
down vote
favorite
A seemingly simple but a challenging quiz,
find the nth term in 1,6,30,120.....
I am thinking is a quadratic due to the constant difference but n th term is needed not a recursive one - I got confused
numerical-methods
asked Nov 15 at 1:30
Waitara Mburu
262
There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A seemingly simple but a challenging quiz,
find the nth term in 1,6,30,120.....
I am thinking is a quadratic due to the constant difference but n th term is needed not a recursive one - I got confused
numerical-methods
asked Nov 15 at 1:30
Waitara Mburu
262
A seemingly simple but a challenging quiz,
find the nth term in 1,6,30,120.....
I am thinking is a quadratic due to the constant difference but n th term is needed not a recursive one - I got confused
numerical-methods
numerical-methods
asked Nov 15 at 1:30
Waitara Mburu
262
asked Nov 15 at 1:30
Waitara Mburu
262
asked Nov 15 at 1:30
Waitara Mburu
262
asked Nov 15 at 1:30
Waitara Mburu
262
asked Nov 15 at 1:30
Waitara Mburu
262
262
There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40
add a comment |
There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40
There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40
add a comment |
2 Answers
2
active
oldest
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up vote
0
down vote
It seems to me like $$ a_1=1$$
$$ a_2=1times 6 =6$$
$$ a_3 = 6times 5 =30$$
$$ a_4 =30times 4 =120$$
$$ a_5 =120times 3 = 360$$
Thus $$a_1=1, a_2=6$$
And for $nge 3$,
$$ a_{n+1} = (8-n)a_n $$
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
add a comment |
up vote
0
down vote
We observe that, 1×6=6
6×5=30
30×4=120
In general,$$a_{n+1} = (7-n)×a_n$$
Divide by $a_n$ on both sides and let that sequence be $b_n$(which is an AP). Solve fore $b_n$ and hence find $a_n$. But you can divide both sides only if $a_n neq 0$ and you don't need to solve if $a_n=0$ as all further terms are 0.
Hope it is helpful
answered Nov 15 at 1:48
Crazy for maths
5309
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It seems to me like $$ a_1=1$$
$$ a_2=1times 6 =6$$
$$ a_3 = 6times 5 =30$$
$$ a_4 =30times 4 =120$$
$$ a_5 =120times 3 = 360$$
Thus $$a_1=1, a_2=6$$
And for $nge 3$,
$$ a_{n+1} = (8-n)a_n $$
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
add a comment |
up vote
0
down vote
It seems to me like $$ a_1=1$$
$$ a_2=1times 6 =6$$
$$ a_3 = 6times 5 =30$$
$$ a_4 =30times 4 =120$$
$$ a_5 =120times 3 = 360$$
Thus $$a_1=1, a_2=6$$
And for $nge 3$,
$$ a_{n+1} = (8-n)a_n $$
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
add a comment |
up vote
0
down vote
up vote
0
down vote
It seems to me like $$ a_1=1$$
$$ a_2=1times 6 =6$$
$$ a_3 = 6times 5 =30$$
$$ a_4 =30times 4 =120$$
$$ a_5 =120times 3 = 360$$
Thus $$a_1=1, a_2=6$$
And for $nge 3$,
$$ a_{n+1} = (8-n)a_n $$
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
It seems to me like $$ a_1=1$$
$$ a_2=1times 6 =6$$
$$ a_3 = 6times 5 =30$$
$$ a_4 =30times 4 =120$$
$$ a_5 =120times 3 = 360$$
Thus $$a_1=1, a_2=6$$
And for $nge 3$,
$$ a_{n+1} = (8-n)a_n $$
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
answered Nov 15 at 1:44
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
add a comment |
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
Thank you guys, a friend thought that a polynomial degree 4 could help but obviously that was not correct as that would mean 'continuity!'
– Waitara Mburu
Nov 15 at 8:52
add a comment |
up vote
0
down vote
We observe that, 1×6=6
6×5=30
30×4=120
In general,$$a_{n+1} = (7-n)×a_n$$
Divide by $a_n$ on both sides and let that sequence be $b_n$(which is an AP). Solve fore $b_n$ and hence find $a_n$. But you can divide both sides only if $a_n neq 0$ and you don't need to solve if $a_n=0$ as all further terms are 0.
Hope it is helpful
answered Nov 15 at 1:48
Crazy for maths
5309
add a comment |
up vote
0
down vote
We observe that, 1×6=6
6×5=30
30×4=120
In general,$$a_{n+1} = (7-n)×a_n$$
Divide by $a_n$ on both sides and let that sequence be $b_n$(which is an AP). Solve fore $b_n$ and hence find $a_n$. But you can divide both sides only if $a_n neq 0$ and you don't need to solve if $a_n=0$ as all further terms are 0.
Hope it is helpful
answered Nov 15 at 1:48
Crazy for maths
5309
add a comment |
up vote
0
down vote
up vote
0
down vote
We observe that, 1×6=6
6×5=30
30×4=120
In general,$$a_{n+1} = (7-n)×a_n$$
Divide by $a_n$ on both sides and let that sequence be $b_n$(which is an AP). Solve fore $b_n$ and hence find $a_n$. But you can divide both sides only if $a_n neq 0$ and you don't need to solve if $a_n=0$ as all further terms are 0.
Hope it is helpful
answered Nov 15 at 1:48
Crazy for maths
5309
We observe that, 1×6=6
6×5=30
30×4=120
In general,$$a_{n+1} = (7-n)×a_n$$
Divide by $a_n$ on both sides and let that sequence be $b_n$(which is an AP). Solve fore $b_n$ and hence find $a_n$. But you can divide both sides only if $a_n neq 0$ and you don't need to solve if $a_n=0$ as all further terms are 0.
Hope it is helpful
answered Nov 15 at 1:48
Crazy for maths
5309
answered Nov 15 at 1:48
Crazy for maths
5309
answered Nov 15 at 1:48
Crazy for maths
5309
answered Nov 15 at 1:48
Crazy for maths
5309
5309
add a comment |
add a comment |
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There are infinitely many ways to continue this. Do you have any additional information?
– Vasya
Nov 15 at 1:39
It seems that the formula is multiplying the previous one by $7-n$ with $a_0=1.$ So, $a_1=1(7-1)=6,$ $a_2=a_1(7-2)=30,$ and so on.
– Will M.
Nov 15 at 1:40
Sequences are not determined by any finite number of terms. We can only guess. This one looks like $$1,quad6,quad6times5,quad6times5times4,quad6times5times4times3,quadcdots$$ It quickly reaches $0$ and stays there.
– mr_e_man
Nov 15 at 1:40