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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?

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Well, $k$ is forwards, $n-k$ is ... –  leo Jan 20 '12 at 3:32
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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
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1 Answer

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.

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I get it now, thanks! –  Jeremy Raymond Jan 20 '12 at 12:30
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