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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
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3

1

I have already aligned the first set of these equations

beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k = S &= 1+2+3+...+n\
 &= n + (n-1) + (n-2) + ... + 1\
 therefore 2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 &= n(n+1)\
 therefore S &= frac12 n (n+1)\
endalign*

However I would like the first equals in the line with two equals to be aligned with the preceding equals signs but all the following equals signs should be aligned with the second of the equals signs in the line with two since obviously that argument follows on.

Is there a way to do this?

horizontal-alignment align

share|improve this question

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  To be honest, I see no reason to align the equals signs after the summations: they aren't related to each other.
  
  – egreg
  Sep 26 at 19:27
  

  
  
  
  

add a comment
 | 

3

1

I have already aligned the first set of these equations

beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k = S &= 1+2+3+...+n\
 &= n + (n-1) + (n-2) + ... + 1\
 therefore 2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 &= n(n+1)\
 therefore S &= frac12 n (n+1)\
endalign*

However I would like the first equals in the line with two equals to be aligned with the preceding equals signs but all the following equals signs should be aligned with the second of the equals signs in the line with two since obviously that argument follows on.

Is there a way to do this?

horizontal-alignment align

share|improve this question

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  To be honest, I see no reason to align the equals signs after the summations: they aren't related to each other.
  
  – egreg
  Sep 26 at 19:27
  

  
  
  
  

add a comment
 | 

I have already aligned the first set of these equations

beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k = S &= 1+2+3+...+n\
 &= n + (n-1) + (n-2) + ... + 1\
 therefore 2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 &= n(n+1)\
 therefore S &= frac12 n (n+1)\
endalign*

However I would like the first equals in the line with two equals to be aligned with the preceding equals signs but all the following equals signs should be aligned with the second of the equals signs in the line with two since obviously that argument follows on.

Is there a way to do this?

horizontal-alignment align

share|improve this question

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k = S &= 1+2+3+...+n\
 &= n + (n-1) + (n-2) + ... + 1\
 therefore 2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 &= n(n+1)\
 therefore S &= frac12 n (n+1)\
endalign*

share|improve this question

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

share|improve this question

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

edited Sep 26 at 19:02

Werner

476k8181 gold badges10791079 silver badges18261826 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

asked Sep 26 at 18:02

Lewis LockwoodLewis Lockwood

3322 bronze badges

3 Answers
3

active

oldest

votes

6

Slightly more complicated than Steven's answer but allows you to add equation numbers and so on if needed. (You can see the difference from the fact that here more space is allocated for the line that has the sum in it.)

documentclassarticle
usepackagemathtools,amssymb,eqparbox
begindocument
beginalign*
sum_k=1^n 1 &= n\
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &eqmakebox[pft]$= S$ = 1+2+3+dots+n\
 &eqmakebox[pft]= n + (n-1) + (n-2) + dots + 1\
 &eqmakebox[pft]mathllaptherefore 2S= (n+1) + (n+1) + (n+1) + dots+ (n+1)\
 &eqmakebox[pft]= n(n+1)\
 &eqmakebox[pft]mathllaptherefore 2S= frac12 n (n+1)\
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:06

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

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add a comment
 | 

7

documentclassarticle
usepackagemathtools,amssymb
begindocument
beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &= beginaligned[t] S &= 1+2+3+dots+n\
 &= n + (n-1) + (n-2) +dots+ 1\
 mathllaptherefore 2S &= (n+1) + (n+1) + (n+1) +dots+ (n+1)\
 &= n(n+1)\
 mathllaptherefore S &= frac12 n (n+1)
endaligned
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:52

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

add a comment
 | 

3

You can have what you want (I think) with alignat*and, in addition the alignment of the therefore symbols. I replaced the fraction 1/2 with a medium-size fraction, which look better, in my opinion.

documentclassarticle
usepackage[utf8]inputenc%
usepackagenccmath 
usepackagemathtools, amssymb


begindocument

beginalignat*2
sum_k=1^n 1 &= n\
sum_k=m^n 1 &=mathrlap n+1-m\
mathopsmash[b]sum_k=1^n k & = S & &= 1+2+3+...+n\
 & & &= n + (n-1) + (n-2) + ... + 1\
 therefore && mathllap2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 & & &= n(n+1)\
 therefore && mathllapS &= mfrac12 n (n+1)\
endalignat*

enddocument

share|improve this answer

edited Sep 26 at 19:41

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mathopsmash[b]...
  
  – egreg
  Sep 26 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @egreg: I didn't know it was required. Upon reflection, though, it's natural: the sum is beteen a pair of braces and therefore becomes mathord. Am I correct?
  
  – Bernard
  Sep 26 at 19:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, that's the problem! You can also, more simply, place k inside the smash.
  
  – egreg
  Sep 26 at 20:09
  
  

  
  
  
  

add a comment
 | 

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6

Slightly more complicated than Steven's answer but allows you to add equation numbers and so on if needed. (You can see the difference from the fact that here more space is allocated for the line that has the sum in it.)

documentclassarticle
usepackagemathtools,amssymb,eqparbox
begindocument
beginalign*
sum_k=1^n 1 &= n\
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &eqmakebox[pft]$= S$ = 1+2+3+dots+n\
 &eqmakebox[pft]= n + (n-1) + (n-2) + dots + 1\
 &eqmakebox[pft]mathllaptherefore 2S= (n+1) + (n+1) + (n+1) + dots+ (n+1)\
 &eqmakebox[pft]= n(n+1)\
 &eqmakebox[pft]mathllaptherefore 2S= frac12 n (n+1)\
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:06

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

add a comment
 | 

6

Slightly more complicated than Steven's answer but allows you to add equation numbers and so on if needed. (You can see the difference from the fact that here more space is allocated for the line that has the sum in it.)

documentclassarticle
usepackagemathtools,amssymb,eqparbox
begindocument
beginalign*
sum_k=1^n 1 &= n\
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &eqmakebox[pft]$= S$ = 1+2+3+dots+n\
 &eqmakebox[pft]= n + (n-1) + (n-2) + dots + 1\
 &eqmakebox[pft]mathllaptherefore 2S= (n+1) + (n+1) + (n+1) + dots+ (n+1)\
 &eqmakebox[pft]= n(n+1)\
 &eqmakebox[pft]mathllaptherefore 2S= frac12 n (n+1)\
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:06

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

add a comment
 | 

Slightly more complicated than Steven's answer but allows you to add equation numbers and so on if needed. (You can see the difference from the fact that here more space is allocated for the line that has the sum in it.)

documentclassarticle
usepackagemathtools,amssymb,eqparbox
begindocument
beginalign*
sum_k=1^n 1 &= n\
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &eqmakebox[pft]$= S$ = 1+2+3+dots+n\
 &eqmakebox[pft]= n + (n-1) + (n-2) + dots + 1\
 &eqmakebox[pft]mathllaptherefore 2S= (n+1) + (n+1) + (n+1) + dots+ (n+1)\
 &eqmakebox[pft]= n(n+1)\
 &eqmakebox[pft]mathllaptherefore 2S= frac12 n (n+1)\
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:06

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

documentclassarticle
usepackagemathtools,amssymb,eqparbox
begindocument
beginalign*
sum_k=1^n 1 &= n\
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &eqmakebox[pft]$= S$ = 1+2+3+dots+n\
 &eqmakebox[pft]= n + (n-1) + (n-2) + dots + 1\
 &eqmakebox[pft]mathllaptherefore 2S= (n+1) + (n+1) + (n+1) + dots+ (n+1)\
 &eqmakebox[pft]= n(n+1)\
 &eqmakebox[pft]mathllaptherefore 2S= frac12 n (n+1)\
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:06

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

answered Sep 26 at 18:17

Schrödinger's catSchrödinger's cat

28.9k22 gold badges4343 silver badges6969 bronze badges

7

documentclassarticle
usepackagemathtools,amssymb
begindocument
beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &= beginaligned[t] S &= 1+2+3+dots+n\
 &= n + (n-1) + (n-2) +dots+ 1\
 mathllaptherefore 2S &= (n+1) + (n+1) + (n+1) +dots+ (n+1)\
 &= n(n+1)\
 mathllaptherefore S &= frac12 n (n+1)
endaligned
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:52

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

add a comment
 | 

7

documentclassarticle
usepackagemathtools,amssymb
begindocument
beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &= beginaligned[t] S &= 1+2+3+dots+n\
 &= n + (n-1) + (n-2) +dots+ 1\
 mathllaptherefore 2S &= (n+1) + (n+1) + (n+1) +dots+ (n+1)\
 &= n(n+1)\
 mathllaptherefore S &= frac12 n (n+1)
endaligned
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:52

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

add a comment
 | 

documentclassarticle
usepackagemathtools,amssymb
begindocument
beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &= beginaligned[t] S &= 1+2+3+dots+n\
 &= n + (n-1) + (n-2) +dots+ 1\
 mathllaptherefore 2S &= (n+1) + (n+1) + (n+1) +dots+ (n+1)\
 &= n(n+1)\
 mathllaptherefore S &= frac12 n (n+1)
endaligned
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:52

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

documentclassarticle
usepackagemathtools,amssymb
begindocument
beginalign*
sum_k=1^n 1 &= n\
newline
sum_k=m^n 1 &= n+1-m\
sum_k=1^n k &= beginaligned[t] S &= 1+2+3+dots+n\
 &= n + (n-1) + (n-2) +dots+ 1\
 mathllaptherefore 2S &= (n+1) + (n+1) + (n+1) +dots+ (n+1)\
 &= n(n+1)\
 mathllaptherefore S &= frac12 n (n+1)
endaligned
endalign*
enddocument

share|improve this answer

edited Sep 26 at 19:52

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

answered Sep 26 at 18:11

Steven B. SegletesSteven B. Segletes

176k99 gold badges224224 silver badges456456 bronze badges

3

You can have what you want (I think) with alignat*and, in addition the alignment of the therefore symbols. I replaced the fraction 1/2 with a medium-size fraction, which look better, in my opinion.

documentclassarticle
usepackage[utf8]inputenc%
usepackagenccmath 
usepackagemathtools, amssymb


begindocument

beginalignat*2
sum_k=1^n 1 &= n\
sum_k=m^n 1 &=mathrlap n+1-m\
mathopsmash[b]sum_k=1^n k & = S & &= 1+2+3+...+n\
 & & &= n + (n-1) + (n-2) + ... + 1\
 therefore && mathllap2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 & & &= n(n+1)\
 therefore && mathllapS &= mfrac12 n (n+1)\
endalignat*

enddocument

share|improve this answer

edited Sep 26 at 19:41

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mathopsmash[b]...
  
  – egreg
  Sep 26 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @egreg: I didn't know it was required. Upon reflection, though, it's natural: the sum is beteen a pair of braces and therefore becomes mathord. Am I correct?
  
  – Bernard
  Sep 26 at 19:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, that's the problem! You can also, more simply, place k inside the smash.
  
  – egreg
  Sep 26 at 20:09
  
  

  
  
  
  

add a comment
 | 

3

You can have what you want (I think) with alignat*and, in addition the alignment of the therefore symbols. I replaced the fraction 1/2 with a medium-size fraction, which look better, in my opinion.

documentclassarticle
usepackage[utf8]inputenc%
usepackagenccmath 
usepackagemathtools, amssymb


begindocument

beginalignat*2
sum_k=1^n 1 &= n\
sum_k=m^n 1 &=mathrlap n+1-m\
mathopsmash[b]sum_k=1^n k & = S & &= 1+2+3+...+n\
 & & &= n + (n-1) + (n-2) + ... + 1\
 therefore && mathllap2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 & & &= n(n+1)\
 therefore && mathllapS &= mfrac12 n (n+1)\
endalignat*

enddocument

share|improve this answer

edited Sep 26 at 19:41

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  mathopsmash[b]...
  
  – egreg
  Sep 26 at 19:31
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @egreg: I didn't know it was required. Upon reflection, though, it's natural: the sum is beteen a pair of braces and therefore becomes mathord. Am I correct?
  
  – Bernard
  Sep 26 at 19:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, that's the problem! You can also, more simply, place k inside the smash.
  
  – egreg
  Sep 26 at 20:09
  
  

  
  
  
  

add a comment
 | 

You can have what you want (I think) with alignat*and, in addition the alignment of the therefore symbols. I replaced the fraction 1/2 with a medium-size fraction, which look better, in my opinion.

documentclassarticle
usepackage[utf8]inputenc%
usepackagenccmath 
usepackagemathtools, amssymb


begindocument

beginalignat*2
sum_k=1^n 1 &= n\
sum_k=m^n 1 &=mathrlap n+1-m\
mathopsmash[b]sum_k=1^n k & = S & &= 1+2+3+...+n\
 & & &= n + (n-1) + (n-2) + ... + 1\
 therefore && mathllap2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 & & &= n(n+1)\
 therefore && mathllapS &= mfrac12 n (n+1)\
endalignat*

enddocument

share|improve this answer

edited Sep 26 at 19:41

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

documentclassarticle
usepackage[utf8]inputenc%
usepackagenccmath 
usepackagemathtools, amssymb


begindocument

beginalignat*2
sum_k=1^n 1 &= n\
sum_k=m^n 1 &=mathrlap n+1-m\
mathopsmash[b]sum_k=1^n k & = S & &= 1+2+3+...+n\
 & & &= n + (n-1) + (n-2) + ... + 1\
 therefore && mathllap2S &= (n+1) + (n+1) + (n+1) + ...+ (n+1)\
 & & &= n(n+1)\
 therefore && mathllapS &= mfrac12 n (n+1)\
endalignat*

enddocument

share|improve this answer

edited Sep 26 at 19:41

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges

answered Sep 26 at 18:58

BernardBernard

199k88 gold badges8888 silver badges234234 bronze badges