My Math Forum hard question with subsets

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 September 24th, 2007, 07:42 AM #1 Newbie   Joined: Sep 2007 Posts: 2 Thanks: 0 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).
 September 24th, 2007, 12:25 PM #2 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 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.
 September 24th, 2007, 02:28 PM #3 Newbie   Joined: Sep 2007 Posts: 2 Thanks: 0 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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