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My question is about the simplification of the following summation: $\sum_{n = 0}^{w}({ \sum_{k = n}^{2n} (k) })$.

I have used Mathematica and obtained that $ \sum_{n = 0}^{w}({ \sum_{k = n}^{2n} (k) }) = \frac{1}{2} w (w + 1) (w + 2)$. Even though this is all I originally needed to know for the problem, I am curious to know how this simplification is actually done because I don't quite understand it. (I think it just surpasses my current mathematical knowledge).

Could anyone explain this to me?

Thanks in advance.

Kiwii
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2 Answers2

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In this case, we can deal with the inside sum separately.

$\sum_{k=n}^{2n}{k} = \sum_{i=0}^{n}{i+n} = \sum_{i=0}^{n}{i} + \sum_{i=0}^{n}{n} = \frac{n(n+1)}{2} + n(n+1) = \frac{3}{2}n(n+1).$

Therefore, the original sum reduces to $\frac{3}{2}\sum_{n=0}^{w}{n^2 + n}$.

Using the equation for the sum of the first $w$ squares, we have:

$$ \sum_{n=0}^{w}{n^2+n} = \frac{w(w+1)(2w+1)}{6} + \frac{w(w+1)}{2}$$

Simplifying, this reduces to $\frac{w(w+1)(w+2)}{3}$.

Bringing back in the leading factor of $3/2$ yields the final result of: $$ \sum_{n=0}^{w}{\sum_{k=n}^{2n}{k}} = \frac{1}{2} w(w+1)(w+2)$$

andars
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  • Thank you! Even though I've already accepted someone else's answer, yours is correct and helpful as well so I've accepted it too! Edit: I didn't because I can't accept two answers? – Kiwii Apr 24 '17 at 20:11
  • Quite welcome. You can only accept one. I recommend @Jennifer's answer; it is more complete. – andars Apr 24 '17 at 20:12
  • Done! And again, thanks. – Kiwii Apr 24 '17 at 20:13
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Using : $$\sum_{k = 0}^{n} k=\frac{n(n+1)}{2}$$ $$\sum_{k = 0}^{n} k^2=\frac{n(n+1)(2n+1)}{6}$$ We get : $$\sum_{n = 0}^{w}{ \sum_{k = n}^{2n} k }=\sum_{n = 0}^{w}\left(\sum_{k = 0}^{2n} k -\sum_{k = 0}^{n-1} k \right)=\sum_{n = 0}^{w}\left(\frac{(2n+1)2n}{2} -\frac{n(n-1)}2 \right)\\=\frac32\sum_{n = 0}^{w}n^2 +\frac32\sum_{n = 0}^{w}n\\=\frac32\left(\frac{w(w+1)(2w+1)}{6}+\frac{w(w+1)}{2}\right)\\=\frac32w(w+1)\frac12\left(\frac{2w+1}{3}+1\right)\\=\frac14w(w+1)(2w+4)\\=\frac12w(w+1)(w+2)$$

Bérénice
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