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I'm trying to figure out the value of this sum, and if I'm correct it appears to be 0.

Check please this solution.

So we have $n-1$ elements ($S_iS_{i+1}$), and every element can be $1$ or $-1$. Because if we put in first element 1 (for example $S_1 = 1, S_2 = 1$) than we can put $1$, or $-1$ in next elements ($S_3 = -1$, or $1$). And for sum of our combination of elements $= k$ we have another combination, which sum $=-k$ (Because number of combinations always $2^n$)

Steven Stadnicki
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Eugene Korotkov
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  • I just wrote a comment on your previous related post - can you explain what is the probability distribution driving the different values or is it deterministic pattern? – Chinny84 Mar 02 '17 at 22:48
  • @Chinny84 Sorry, i'm really bad in english. Cant understand, can you please say it more easier for understanding? Sorry. – Eugene Korotkov Mar 02 '17 at 22:51
  • No worries - how do we get $S_i$? – Chinny84 Mar 02 '17 at 22:54
  • The value of the sum depends on the choice of $S_l$'s (for example you will not have value 0 by taking all $S_l$'s equal 1). Do you ask for the set of all possible values of this sum or what? This question is unclear. – larry01 Mar 02 '17 at 22:54
  • @Chinny84 We need go for all combinations, i mean is $n = 3$, we will go sum of $SiS$j for (1,1,1),(-1,-1,-1),(1,-1,-1),(-1,-1,1) and next ($2^3$ combinations) – Eugene Korotkov Mar 02 '17 at 22:59
  • @Chinny84 larry01 is right, i mean all possible values. Can you edit my question please? Dont know how write it correctly. – Eugene Korotkov Mar 02 '17 at 23:01
  • I've edited the question and inserted a notation that should make it fairly clear. – Steven Stadnicki Mar 02 '17 at 23:45

1 Answers1

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Let $T_i$ denote the sum of $S_iS_{i+1}$ over $\left(S_1,S_2,\ldots,S_n\right)\in\{-1,+1\}^n$. Clearly, the required sum is $\sum_{i=1}^{n-1}\,T_i$. You only have to show that $T_i=0$ for all $i$, but this is quite easy to prove.

Batominovski
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