Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In a math text I have an equation:

S = $\sum \limits_{0 \leq k \leq n} (a + bk)$

By the communitive law we can replace $k$ by $n - k$, obtaining

S = $\sum \limits_{0 \leq n-k \leq n} (a + b(n - k)) = \sum \limits_{0 \leq k \leq n}(a + bn -bk))$

In the last equation how does the index $0 \leq n - k \leq n$ become $0 \leq k \leq n$ again?

share|cite|improve this question
Well, $k$ is forwards, $n-k$ is ... – leo Jan 20 '12 at 3:32
When $n-k\leq n$ then $0 \leq k,$ similarly, when $0\leq n-k$, then $k\leq n.$ – Ehsan M. Kermani Jan 20 '12 at 3:35
Try with some particular examples, take $n=3, 4$ – leo Jan 20 '12 at 3:35
up vote 2 down vote accepted

In the first case, $n-k$ decreases like $n,n-1,n-2,\dots,0$ whereas the second case the terms are ordered $0,1,\dots,n$. Thus the re-indexing corresponds to reordering the sum. If addition is commutative, this causes no problem and we have equality.

share|cite|improve this answer
I get it now, thanks! – Jeremy Raymond Jan 20 '12 at 12:30

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.