1025th term of the given sequence.Finding the nth term in a repeating number sequenceHow to show all the terms in a sequence are greater than a number?Find first term and common difference of Arithmetic Sequence, given two other termsThe second term of an arithmetic sequence is $13$ and $5^th$ term is $31$. What is the $17^th$ term of the sequence?Unique sequenceHow to calculate the nth term of sequence that increases by n?Finding arithmetic sequence first termGeometric Sequence with formula for kth term.Description of an Increasing Maximum Value in a Sequence of IntegersWhat is the general term of this sequence of integers?
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1025th term of the given sequence.
Finding the nth term in a repeating number sequenceHow to show all the terms in a sequence are greater than a number?Find first term and common difference of Arithmetic Sequence, given two other termsThe second term of an arithmetic sequence is $13$ and $5^th$ term is $31$. What is the $17^th$ term of the sequence?Unique sequenceHow to calculate the nth term of sequence that increases by n?Finding arithmetic sequence first termGeometric Sequence with formula for kth term.Description of an Increasing Maximum Value in a Sequence of IntegersWhat is the general term of this sequence of integers?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^th, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
1 - 1
2 - 2
3 - 2
4 - 4
5 - 4
. . .
8 - 8
9 - 8
. . .
We can notice that $ 4^th$ term is 4 and similarly, the $ 8^th$ term is 8.
So the $ 1025^th$ term must be 1024 as $ 1024^th $ term starts with 1024.
So the value of $ 1025^th$ term is $ 2^10 $ .
Is there any other method to solve this question?
sequences-and-series
asked 9 hours ago
KaushikKaushik
716 bronze badges
$endgroup$
add a comment |
$begingroup$
Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^th, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
1 - 1
2 - 2
3 - 2
4 - 4
5 - 4
. . .
8 - 8
9 - 8
. . .
We can notice that $ 4^th$ term is 4 and similarly, the $ 8^th$ term is 8.
So the $ 1025^th$ term must be 1024 as $ 1024^th $ term starts with 1024.
So the value of $ 1025^th$ term is $ 2^10 $ .
Is there any other method to solve this question?
sequences-and-series
asked 9 hours ago
KaushikKaushik
716 bronze badges
$endgroup$
1
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago
add a comment |
$begingroup$
Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^th, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
1 - 1
2 - 2
3 - 2
4 - 4
5 - 4
. . .
8 - 8
9 - 8
. . .
We can notice that $ 4^th$ term is 4 and similarly, the $ 8^th$ term is 8.
So the $ 1025^th$ term must be 1024 as $ 1024^th $ term starts with 1024.
So the value of $ 1025^th$ term is $ 2^10 $ .
Is there any other method to solve this question?
sequences-and-series
asked 9 hours ago
KaushikKaushik
716 bronze badges
$endgroup$
Consider the following sequence - $$ 1,2,2,4,4,4,4,8,8,8,8,8,8,8,8, ... $$
In this sequence, what will be the $ 1025^th, term $
So, when we write down the sequence and then write the value of $ n $ (Here, $n$ stands for the number of the below term) above it We can observe the following -
1 - 1
2 - 2
3 - 2
4 - 4
5 - 4
. . .
8 - 8
9 - 8
. . .
We can notice that $ 4^th$ term is 4 and similarly, the $ 8^th$ term is 8.
So the $ 1025^th$ term must be 1024 as $ 1024^th $ term starts with 1024.
So the value of $ 1025^th$ term is $ 2^10 $ .
Is there any other method to solve this question?
sequences-and-series
sequences-and-series
asked 9 hours ago
KaushikKaushik
716 bronze badges
asked 9 hours ago
KaushikKaushik
716 bronze badges
asked 9 hours ago
KaushikKaushik
716 bronze badges
asked 9 hours ago
KaushikKaushik
716 bronze badges
asked 9 hours ago
KaushikKaushik
716 bronze badges
716 bronze badges
1
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago
add a comment |
1
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago
1
1
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
A higher brow way of writing the same thing is to say the $n^th$ term is
$2^lfloor log_2 nrfloor$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^10=1024$.
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
$endgroup$
add a comment |
$begingroup$
In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,cdots$$
So for any term, clear all bits but the most significant.
$$10000000001to10000000000.$$
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
A higher brow way of writing the same thing is to say the $n^th$ term is
$2^lfloor log_2 nrfloor$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^10=1024$.
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
$endgroup$
add a comment |
$begingroup$
A higher brow way of writing the same thing is to say the $n^th$ term is
$2^lfloor log_2 nrfloor$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^10=1024$.
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
$endgroup$
add a comment |
$begingroup$
A higher brow way of writing the same thing is to say the $n^th$ term is
$2^lfloor log_2 nrfloor$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^10=1024$.
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
$endgroup$
A higher brow way of writing the same thing is to say the $n^th$ term is
$2^lfloor log_2 nrfloor$, then to evaluate that at $n=1025$. The base $2$ log of $1025$ is slightly greater than $10$, so the term is $2^10=1024$.
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
answered 9 hours ago
Ross MillikanRoss Millikan
309k24 gold badges203 silver badges381 bronze badges
309k24 gold badges203 silver badges381 bronze badges
add a comment |
add a comment |
$begingroup$
In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,cdots$$
So for any term, clear all bits but the most significant.
$$10000000001to10000000000.$$
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
$endgroup$
add a comment |
$begingroup$
In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,cdots$$
So for any term, clear all bits but the most significant.
$$10000000001to10000000000.$$
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
$endgroup$
add a comment |
$begingroup$
In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,cdots$$
So for any term, clear all bits but the most significant.
$$10000000001to10000000000.$$
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
$endgroup$
In binary, the term indexes
$$1,10,11,100,101,110,111,1000,1001,1010,1011,1100,1101,1110,1111,cdots$$
become
$$1,10,10,100,100,100,100,1000,1000,1000,1000,1000,1000,1000,1000,cdots$$
So for any term, clear all bits but the most significant.
$$10000000001to10000000000.$$
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
edited 8 hours ago
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
answered 8 hours ago
Yves DaoustYves Daoust
142k9 gold badges85 silver badges241 bronze badges
142k9 gold badges85 silver badges241 bronze badges
add a comment |
add a comment |
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1
$begingroup$
This method is very efficient. Why would you want another ?
$endgroup$
– Yves Daoust
8 hours ago