Does $$\sum_{r=1}^n(ar+b)=\sum_{r=1}^nar+b$$ or does $$\sum_{r=1}^n(ar+b)=\sum_{r=1}^nar+\sum_{r=1}^nb$$ If I'm given $u_r=ar+b$ how would I substitute that into $$\sum_{r=1}^nu_r$$ Does that mean $$\sum_{r=1}^nu_r=\sum_{r=1}^n(u_r)=\sum_{r=1}^n(ar+b)$$ or $$\sum_{r=1}^nu_r=\sum_{r=1}^nar+b$$ Also $u_r$ means $f(r)$, right?
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1The latter is correct. You can regard $u_r$ as $f(r)$. – string Apr 12 '15 at 08:49
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You can see the correct answer by writing the sums explicitly: $$\begin{align} \sum_{r=1}^n(ar+b) &= (a1+b) + (a2+b) + \dots + (an+b)\\ &=(a1 + a2 + \dots + an) + (b + b + \dots + b)\\ &=\sum_{r=1}^nar+\sum_{r=1}^nb \end{align}$$
celtschk
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1Nice way of showing clearly why the latter equation is correct. +1 – Daniel W. Farlow Apr 12 '15 at 09:36
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This seems to be based on the meaning of the parentheses, in terms of notation. Yes, when you remove the parentheses, you should distribute the "sigma" to both the term ar AND the term b.
Tom
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There is an ambiguity in the notation
$$\sum_{k=1}^n ak+b $$
does it means
$$\left(\sum_{k=1}^n ak\right)+b $$
or
$$\sum_{k=1}^n \left( ak+b \right)$$
Tryss
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I don't think there's any ambiguity. If I saw the first expression in a text or a journal or other piece of literature, then I would certainly understand it to be the former. If the author meant the latter, then that person should not have their work published or the editor should be fired. Context usually makes it clear but the first "interpretation" is the correct one. – Daniel W. Farlow Apr 12 '15 at 09:42