$$\sum_{i=1}^{n}f(i)+k = \sum_{i=1}^{n}\{f(i)+k\} = k+\sum_{i=1}^{n}f(i) $$
I'm more confused about the working of expression: $$\sum_{i=1}^{n}\{f(i)+k\}$$
Are all the 3 expressions equivalent?
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In the first expression put the parentheses to make it more easily understandable: is it $$\sum_{i=1}^n f(i)+k = \sum_{i=1}^n (f(i)+k)? $$ – Paolo Franchi Jan 13 '16 at 09:27
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Nope. But these are:$$\sum_{i=1}^{n}f(i)+nk = \sum_{i=1}^{n}\{f(i)+k\} = nk+\sum_{i=1}^{n}f(i) $$
If you're summing a constant $n$ times, you need to multiply it by $n$ if you want to bring it outside the summation
Ant
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None of the above expressions in the question are equivalent? – Siddharth Thevaril Jan 13 '16 at 09:24
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@Sidsec9 No, because there is no $n$ in front of $k$. I mean you can try it yourself, take $f(i) = i$, $n=2$, $k=3$ and see if the results are the same – Ant Jan 13 '16 at 09:25
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The second and third are different. The second one is
$$\{f(1)+k\}+\{f(2)+k\}+\{f(3)+k\}+...$$ which contains $n$ lots of $k$, so it equals $kn+\sum_{i=1}^nf(i)$.
The first one is ambiguous; I am not certain whether the sum includes $k$ within its scope, so it is equal to either the second or third one.
Empy2
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I understand what you do there, but I think it's better to avoid writing a sum of sets: ${a}+{b}$ is neither $a+b$ nor ${a+b}$ (though there are of course canonical bijections between these objects) – Paolo Franchi Jan 13 '16 at 09:33