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September 24th, 2007, 07:42 AM   #1
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hard question with subsets

We have n>=1 and for the every proper subset A of the set {1,2,...,n} the number w(A) is given as follows: If a1>a2>a3>...>ak are the all elements of set A, then w(A)=a1-a2+a3-...+(-1)^(k+1)*ak.

Calculate the sum of all 2^(n-1) integers w(A).
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September 24th, 2007, 12:25 PM   #2
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W({1}) = 1

W({1, 2}) = 1
W({2}) = 2

W({1, 2, 3}) = 2
W({2, 3}) = 1
W({1, 3}) = 2
W({3}) = 3

W({1, 2, 3, 4}) = 2
W({2, 3, 4}) = 3
W({1, 3, 4}) = 2
W({1, 2, 4}) = 3
W({3, 4}) = 1
W({2, 4}) = 2
W({1, 4}) = 3
W({4}) = 4


Just thinking, don't have anything yet.
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September 24th, 2007, 02:28 PM   #3
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It seems for me that general solution will be S(w(A))=n2^(n-1).
For n=1 we have one subset S(w(A))=1*2^0=1.
For n=2 three subsets and S(w(a))=4=2*2^1.
For n=3 seven subsets and S(w(a))=1+1+2+2+1+2+3=12=3*2^(3-1).
For n=4 fifteen subsets and S(w(a))=4+12+9+6+1=32=4*2^(4-1).

And for n=5 this sum should be equal 5*2^(5-1)=80.

Now I am trying to prove the above conjecture...
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