What will be the next term in this mathematical sequence?
$begingroup$
What will be next in this series?
$$0, 6, 24, 60, 120, 210, ...$$
I've tried it and noticed that the numbers are multiples of six. But I couldn't make a relation between them.
mathematics pattern calculation-puzzle
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
$endgroup$
add a comment |
$begingroup$
What will be next in this series?
$$0, 6, 24, 60, 120, 210, ...$$
I've tried it and noticed that the numbers are multiples of six. But I couldn't make a relation between them.
mathematics pattern calculation-puzzle
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
$endgroup$
$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32
add a comment |
$begingroup$
What will be next in this series?
$$0, 6, 24, 60, 120, 210, ...$$
I've tried it and noticed that the numbers are multiples of six. But I couldn't make a relation between them.
mathematics pattern calculation-puzzle
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
$endgroup$
What will be next in this series?
$$0, 6, 24, 60, 120, 210, ...$$
I've tried it and noticed that the numbers are multiples of six. But I couldn't make a relation between them.
mathematics pattern calculation-puzzle
mathematics pattern calculation-puzzle
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
edited Dec 25 '18 at 9:46
Rand al'Thor
71.1k14235471
71.1k14235471
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
asked Dec 25 '18 at 9:36
Gurbir SinghGurbir Singh
1193
1193
$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32
add a comment |
$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32
$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32
$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
It seems like
sum of digit series
f(n)=n*sumOfDigits(n-1)*sumOfDigits(n+1)
example:-
f(1)=1*sumOfDigits(1-1)*sumOfDigits(1+1) = 1*0*2 = 0
f(2)=2*sumOfDigits(2-1)*sumOfDigits(2+1) = 2*1*3 = 6
f(3)=3*sumOfDigits(3-1)*sumOfDigits(3+1) = 3*2*4 = 24
f(4)=4*sumOfDigits(4-1)*sumOfDigits(4+1) = 4*3*5 = 60
f(5)=5*sumOfDigits(5-1)*sumOfDigits(5+1) = 5*4*6 = 120
f(6)=6*sumOfDigits(6-1)*sumOfDigits(6+1) = 6*5*7 = 210
Answer:-
f(7)=7*sumOfDigits(7-1)*sumOfDigits(7+1) = 7*6*8 = 336
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
$endgroup$
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
add a comment |
$begingroup$
It seems like a
cube series.
Namely,
cube of 1 is 1 then 1 - 1 = 0
cube of 2 is 8 then 8 - 2 = 6
cube of 3 is 27 then 27 - 3 = 24
cube of 4 is 64 then 64 - 4 = 60
cube of 5 is 125 then 125 - 5 = 120
cube of 6 is 216 then 216 - 6 = 210
cube of 7 is 343 then 343 - 7 = 336
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
$endgroup$
add a comment |
$begingroup$
Take differences between terms:
$6, 18, 36, 60, 90, ...$
Notice that these are
$6$ times the triangular numbers $1, 3, 6, 12, 15, ...$
So the next difference should be
$6times 21 = 126$
and the next term should be
$210+126=336$.
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
$endgroup$
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
add a comment |
$begingroup$
Simple Answer:
The $n$th term is $n(n-1)(n+1)$ and thus the required one is the seventh term giving an answer of $$7(6)(8)=336.$$
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
add a comment |
Your Answer
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It seems like
sum of digit series
f(n)=n*sumOfDigits(n-1)*sumOfDigits(n+1)
example:-
f(1)=1*sumOfDigits(1-1)*sumOfDigits(1+1) = 1*0*2 = 0
f(2)=2*sumOfDigits(2-1)*sumOfDigits(2+1) = 2*1*3 = 6
f(3)=3*sumOfDigits(3-1)*sumOfDigits(3+1) = 3*2*4 = 24
f(4)=4*sumOfDigits(4-1)*sumOfDigits(4+1) = 4*3*5 = 60
f(5)=5*sumOfDigits(5-1)*sumOfDigits(5+1) = 5*4*6 = 120
f(6)=6*sumOfDigits(6-1)*sumOfDigits(6+1) = 6*5*7 = 210
Answer:-
f(7)=7*sumOfDigits(7-1)*sumOfDigits(7+1) = 7*6*8 = 336
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
$endgroup$
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
add a comment |
$begingroup$
It seems like
sum of digit series
f(n)=n*sumOfDigits(n-1)*sumOfDigits(n+1)
example:-
f(1)=1*sumOfDigits(1-1)*sumOfDigits(1+1) = 1*0*2 = 0
f(2)=2*sumOfDigits(2-1)*sumOfDigits(2+1) = 2*1*3 = 6
f(3)=3*sumOfDigits(3-1)*sumOfDigits(3+1) = 3*2*4 = 24
f(4)=4*sumOfDigits(4-1)*sumOfDigits(4+1) = 4*3*5 = 60
f(5)=5*sumOfDigits(5-1)*sumOfDigits(5+1) = 5*4*6 = 120
f(6)=6*sumOfDigits(6-1)*sumOfDigits(6+1) = 6*5*7 = 210
Answer:-
f(7)=7*sumOfDigits(7-1)*sumOfDigits(7+1) = 7*6*8 = 336
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
$endgroup$
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
add a comment |
$begingroup$
It seems like
sum of digit series
f(n)=n*sumOfDigits(n-1)*sumOfDigits(n+1)
example:-
f(1)=1*sumOfDigits(1-1)*sumOfDigits(1+1) = 1*0*2 = 0
f(2)=2*sumOfDigits(2-1)*sumOfDigits(2+1) = 2*1*3 = 6
f(3)=3*sumOfDigits(3-1)*sumOfDigits(3+1) = 3*2*4 = 24
f(4)=4*sumOfDigits(4-1)*sumOfDigits(4+1) = 4*3*5 = 60
f(5)=5*sumOfDigits(5-1)*sumOfDigits(5+1) = 5*4*6 = 120
f(6)=6*sumOfDigits(6-1)*sumOfDigits(6+1) = 6*5*7 = 210
Answer:-
f(7)=7*sumOfDigits(7-1)*sumOfDigits(7+1) = 7*6*8 = 336
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
$endgroup$
It seems like
sum of digit series
f(n)=n*sumOfDigits(n-1)*sumOfDigits(n+1)
example:-
f(1)=1*sumOfDigits(1-1)*sumOfDigits(1+1) = 1*0*2 = 0
f(2)=2*sumOfDigits(2-1)*sumOfDigits(2+1) = 2*1*3 = 6
f(3)=3*sumOfDigits(3-1)*sumOfDigits(3+1) = 3*2*4 = 24
f(4)=4*sumOfDigits(4-1)*sumOfDigits(4+1) = 4*3*5 = 60
f(5)=5*sumOfDigits(5-1)*sumOfDigits(5+1) = 5*4*6 = 120
f(6)=6*sumOfDigits(6-1)*sumOfDigits(6+1) = 6*5*7 = 210
Answer:-
f(7)=7*sumOfDigits(7-1)*sumOfDigits(7+1) = 7*6*8 = 336
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
answered Dec 27 '18 at 13:03
Vivek KundariyaVivek Kundariya
263
263
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
add a comment |
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
2
2
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
$begingroup$
That's copying my solution
$endgroup$
– TheSimpliFire
Dec 28 '18 at 13:27
add a comment |
$begingroup$
It seems like a
cube series.
Namely,
cube of 1 is 1 then 1 - 1 = 0
cube of 2 is 8 then 8 - 2 = 6
cube of 3 is 27 then 27 - 3 = 24
cube of 4 is 64 then 64 - 4 = 60
cube of 5 is 125 then 125 - 5 = 120
cube of 6 is 216 then 216 - 6 = 210
cube of 7 is 343 then 343 - 7 = 336
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
$endgroup$
add a comment |
$begingroup$
It seems like a
cube series.
Namely,
cube of 1 is 1 then 1 - 1 = 0
cube of 2 is 8 then 8 - 2 = 6
cube of 3 is 27 then 27 - 3 = 24
cube of 4 is 64 then 64 - 4 = 60
cube of 5 is 125 then 125 - 5 = 120
cube of 6 is 216 then 216 - 6 = 210
cube of 7 is 343 then 343 - 7 = 336
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
$endgroup$
add a comment |
$begingroup$
It seems like a
cube series.
Namely,
cube of 1 is 1 then 1 - 1 = 0
cube of 2 is 8 then 8 - 2 = 6
cube of 3 is 27 then 27 - 3 = 24
cube of 4 is 64 then 64 - 4 = 60
cube of 5 is 125 then 125 - 5 = 120
cube of 6 is 216 then 216 - 6 = 210
cube of 7 is 343 then 343 - 7 = 336
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
$endgroup$
It seems like a
cube series.
Namely,
cube of 1 is 1 then 1 - 1 = 0
cube of 2 is 8 then 8 - 2 = 6
cube of 3 is 27 then 27 - 3 = 24
cube of 4 is 64 then 64 - 4 = 60
cube of 5 is 125 then 125 - 5 = 120
cube of 6 is 216 then 216 - 6 = 210
cube of 7 is 343 then 343 - 7 = 336
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
edited Dec 25 '18 at 13:40
Omega Krypton
5,4442847
5,4442847
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
answered Dec 25 '18 at 11:11
Shailendra SharmaShailendra Sharma
1803
1803
add a comment |
add a comment |
$begingroup$
Take differences between terms:
$6, 18, 36, 60, 90, ...$
Notice that these are
$6$ times the triangular numbers $1, 3, 6, 12, 15, ...$
So the next difference should be
$6times 21 = 126$
and the next term should be
$210+126=336$.
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
$endgroup$
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
add a comment |
$begingroup$
Take differences between terms:
$6, 18, 36, 60, 90, ...$
Notice that these are
$6$ times the triangular numbers $1, 3, 6, 12, 15, ...$
So the next difference should be
$6times 21 = 126$
and the next term should be
$210+126=336$.
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
$endgroup$
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
add a comment |
$begingroup$
Take differences between terms:
$6, 18, 36, 60, 90, ...$
Notice that these are
$6$ times the triangular numbers $1, 3, 6, 12, 15, ...$
So the next difference should be
$6times 21 = 126$
and the next term should be
$210+126=336$.
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
$endgroup$
Take differences between terms:
$6, 18, 36, 60, 90, ...$
Notice that these are
$6$ times the triangular numbers $1, 3, 6, 12, 15, ...$
So the next difference should be
$6times 21 = 126$
and the next term should be
$210+126=336$.
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
edited Dec 25 '18 at 12:10
Omega Krypton
5,4442847
5,4442847
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
answered Dec 25 '18 at 9:46
Rand al'ThorRand al'Thor
71.1k14235471
71.1k14235471
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
add a comment |
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
2
2
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
$begingroup$
The multiplication on the third spoiler is incorrect. And also, the sequence can be found on the OEIS as a nice formula: oeis.org/A007531
$endgroup$
– athin
Dec 25 '18 at 11:07
add a comment |
$begingroup$
Simple Answer:
The $n$th term is $n(n-1)(n+1)$ and thus the required one is the seventh term giving an answer of $$7(6)(8)=336.$$
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
add a comment |
$begingroup$
Simple Answer:
The $n$th term is $n(n-1)(n+1)$ and thus the required one is the seventh term giving an answer of $$7(6)(8)=336.$$
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
add a comment |
$begingroup$
Simple Answer:
The $n$th term is $n(n-1)(n+1)$ and thus the required one is the seventh term giving an answer of $$7(6)(8)=336.$$
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
$endgroup$
Simple Answer:
The $n$th term is $n(n-1)(n+1)$ and thus the required one is the seventh term giving an answer of $$7(6)(8)=336.$$
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
answered Dec 25 '18 at 13:45
TheSimpliFireTheSimpliFire
2,155532
2,155532
add a comment |
add a comment |
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$begingroup$
the OEIS has three potential answers to this question.
$endgroup$
– Don Thousand
Dec 25 '18 at 16:32