Compute $S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor$

3

$begingroup$

I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$

But I am not sure if this is correct. Perhaps someone could give an indication.

sequences-and-series number-theory

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write $t$ in terms of $m$ in each case.
  $endgroup$
  – Hello_World
  Jan 11 at 8:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Hello_World Actuallly, you can.
  $endgroup$
  – 5xum
  Jan 11 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asking me to write them in terms of $m$? I can do that if it helps.
  $endgroup$
  – Hello_World
  Jan 11 at 8:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KemonoChen, please write that as an answer so that we can downvote it.
  $endgroup$
  – Carsten S
  Jan 11 at 14:42
  

  
  
  
  

add a comment | 

3

$begingroup$

I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$

But I am not sure if this is correct. Perhaps someone could give an indication.

sequences-and-series number-theory

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write $t$ in terms of $m$ in each case.
  $endgroup$
  – Hello_World
  Jan 11 at 8:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Hello_World Actuallly, you can.
  $endgroup$
  – 5xum
  Jan 11 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asking me to write them in terms of $m$? I can do that if it helps.
  $endgroup$
  – Hello_World
  Jan 11 at 8:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KemonoChen, please write that as an answer so that we can downvote it.
  $endgroup$
  – Carsten S
  Jan 11 at 14:42
  

  
  
  
  

add a comment | 

3

3

3

$begingroup$

I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$

But I am not sure if this is correct. Perhaps someone could give an indication.

sequences-and-series number-theory

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

$endgroup$

I want to compute the following sum
$$S = sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor.$$
Here is what I tried:
$$ S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor.$$
If $m= 2t$ then
$$S =sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{(t-1)t}{2} = t^2.$$
If $m= 2t+1$ then
$$S = sum_{kgeq 0, 2|k}^{m} leftlfloor frac{k}{2}rightrfloor + sum_{kgeq 0, 2not |k}^{m} leftlfloor frac{k}{2}rightrfloor = frac{t(t+1)}{2} + frac{t(t+1)}{2}= t(t+1).$$

But I am not sure if this is correct. Perhaps someone could give an indication.

sequences-and-series number-theory

sequences-and-series number-theory

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

share|cite|improve this question

share|cite|improve this question

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

edited Jan 11 at 15:27

Asaf Karagila♦

304k32430763

304k32430763

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

asked Jan 11 at 8:29

Hello_WorldHello_World

4,12621731

4,12621731

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write $t$ in terms of $m$ in each case.
  $endgroup$
  – Hello_World
  Jan 11 at 8:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Hello_World Actuallly, you can.
  $endgroup$
  – 5xum
  Jan 11 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asking me to write them in terms of $m$? I can do that if it helps.
  $endgroup$
  – Hello_World
  Jan 11 at 8:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KemonoChen, please write that as an answer so that we can downvote it.
  $endgroup$
  – Carsten S
  Jan 11 at 14:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You can write $t$ in terms of $m$ in each case.
  $endgroup$
  – Hello_World
  Jan 11 at 8:35
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Hello_World Actuallly, you can.
  $endgroup$
  – 5xum
  Jan 11 at 8:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you asking me to write them in terms of $m$? I can do that if it helps.
  $endgroup$
  – Hello_World
  Jan 11 at 8:39
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @KemonoChen, please write that as an answer so that we can downvote it.
  $endgroup$
  – Carsten S
  Jan 11 at 14:42
  

  
  
  
  

1

1

$begingroup$
You can write $t$ in terms of $m$ in each case.
$endgroup$
– Hello_World
Jan 11 at 8:35

$begingroup$
You can write $t$ in terms of $m$ in each case.
$endgroup$
– Hello_World
Jan 11 at 8:35

1

1

$begingroup$
@Hello_World Actuallly, you can.
$endgroup$
– 5xum
Jan 11 at 8:38

$begingroup$
@Hello_World Actuallly, you can.
$endgroup$
– 5xum
Jan 11 at 8:38

$begingroup$
Are you asking me to write them in terms of $m$? I can do that if it helps.
$endgroup$
– Hello_World
Jan 11 at 8:39

$begingroup$
Are you asking me to write them in terms of $m$? I can do that if it helps.
$endgroup$
– Hello_World
Jan 11 at 8:39

$begingroup$
@KemonoChen, please write that as an answer so that we can downvote it.
$endgroup$
– Carsten S
Jan 11 at 14:42

$begingroup$
@KemonoChen, please write that as an answer so that we can downvote it.
$endgroup$
– Carsten S
Jan 11 at 14:42

add a comment | 

1 Answer
1

active

oldest

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8

$begingroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$

share|cite|improve this answer

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

$endgroup$

add a comment | 

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1 Answer
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8

$begingroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$

share|cite|improve this answer

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

$endgroup$

add a comment | 

8

$begingroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$

share|cite|improve this answer

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

$endgroup$

add a comment | 

8

8

8

$begingroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$

share|cite|improve this answer

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

$endgroup$

Yes, you are correct. You may also write the result as a more compact formula:
$$sum_{k=0}^{m} leftlfloor frac{k}{2}rightrfloor=
begin{cases}
t^2&text {if $m=2t$}\
t(t+1)&text {if $m=2t+1$}\
end{cases}=leftlfloor frac{m^2}{4}rightrfloor.$$
Indeed, if $m=2t$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2rightrfloor=t^2$$
and if $m=2t+1$ then
$$leftlfloor frac{m^2}{4}rightrfloor=leftlfloor t^2+t+frac{1}{4}rightrfloor=t(t+1).$$

share|cite|improve this answer

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

share|cite|improve this answer

share|cite|improve this answer

edited Jan 11 at 8:46

edited Jan 11 at 8:46

edited Jan 11 at 8:46

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

answered Jan 11 at 8:40

Robert ZRobert Z

96.5k1065136

96.5k1065136

add a comment | 

add a comment | 

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