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Let a sequence $X = \{x_1, x_2, ..., x_n\}$ and $n = k \times l$ in which $k,l \in \mathbb{N}$. We have the following conditions:

$$A)\sum_{i=1}^{n} x_i = 0$$ $$B)\sum_{i=1}^{n} x_i^2 = 1$$

I was wondering if we could convert the following expression to a formula based on $n, k, l$ or anything else:

$$\sum_{i=1}^{l} (x_{k(i-1)+1} + x_{k(i-1)+2} + ... + x_{k(i-1)+k})^2$$ $i.e,$ in a simple word:

$(x_1 + x_2 + ... + x_k)^2 + (x_{k+1} + x_{k+2} + ... + x_{2k})^2 + ... + (x_{k(l-1)+1} + x_{k(l-1)+2} + ... + x_n)^2$

In general, I think it is not possible. But by having the conditions $A$ and $B$, I am not sure whether it is possible or not.

Patris
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  • This comes down to $1 + 2\sum_{i=1}^l\sum_{1\leq j < m \leq k} x_{k(i-1)+j}x_{k(i-1)+m}$ if that's of any use – Lukas Rollier Jan 07 '20 at 15:58
  • Although I didn't use condition $A$, so there might be something easier – Lukas Rollier Jan 07 '20 at 15:59
  • Yes, I know if we expand it, we get what you said, but actually I am wondering if we can somehow convert the multiplication part you mentioned to a better expression. @Interestedstudent – Patris Jan 07 '20 at 16:02
  • Well, I think the problem there is that your sum doesn't really factor nicely. Any element only appears in it's proper 'subproduct' – Lukas Rollier Jan 07 '20 at 16:05
  • Is there a specific reason why you're looking for a simplification of this sum? – Lukas Rollier Jan 07 '20 at 16:05
  • I am just trying to find a simpler version of this sum since I think there should be. @LukasRollier – Patris Jan 07 '20 at 16:38

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