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If I have: $$\sum_{k=1}^n \sum_{i=1}^k(2k)$$

Why does the inner sum become $k$? I can see that it is you can "pull out" the $2k$ and it would leave a $1$, but why can we pull this out? Is it because the lower bound doesn't match the bound in the expression?

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    It’s because within the inner summation $k$ is a constant. In that respect it’s no different from $\sum_{k=1}^n\sum_{i=1}^k(2a)$ where $a=5$, say. – Brian M. Scott Oct 06 '20 at 23:54
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    \begin{eqnarray} \sum_{k=1}^n \sum_{i=1}^k 2k &=& 2 \sum_{k=1}^n k \sum_{i=1}^k 1 \ &=& 2 \sum_{k=1}^n k \underbrace{\sum_{i=1}^k 1}_{=k} \ \end{eqnarray} – Donald Splutterwit Oct 06 '20 at 23:56

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This is just distribution of common factors in a sum . $(ab+ac)=a(b+c)$

You can "pull out" (distribute) factors in a series that do not contain the bound variable (or iterator) for that series (as a term or an index or such).   The domain over which the bound variable is iterated is not a concern.

The bound variable of the inner series is $i$, and $2k$ does not contain $i$.

The distribution of the inner series is thus : $(2k+2k+\ldots+2k)=2k(1+1+\ldots+1)$.$$\sum_{i=1}^k 2k=2k\sum_{i=1}^k1$$

And hence $$\qquad\qquad\begin{align}\sum_{k=1}^n\sum_{i=1}^k 2k&=2\sum_{k=1}^n k^2\\[1ex]&\phantom{=\dfrac 26\cdot n(n+1)(2n+1)}\end{align}$$

Graham Kemp
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