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Let f(n) be the sum of all the 'alternating sums' of{1,2,3.....n} 's subsets.
To evaluatef(n), we can calculate the 'alternating sums' by separating the subsets into two types: contains  and does not contain .

For those 2^(n-1) subsets which does not contain , the sum of 'alternating sums' will be f(n-1) .
For the remaining2^(n-1)  subsets containing ,  must be the greatest number in the subsets.
Every subsets of these can be expressed as {n}聯集A, where A包含在{1,2,3.....n-1}, and so its 'alternating sum' must equal  minus the 'alternating sum' of A .
Therefore the sum of 'alternating sums' of these subsets= sigma[A包含在(1,2,.....n-1)] [n-alternating sum(A)]=2^(n-1) *n-f(n-1).
Summing up, we get f(n)=2^(n-1)n .

1.先求{1}的所有子集合交錯和的總和＝1
{ }＝0
{1}＝1

2.先求{1,2}的所有子集合交錯和的總和＝4
{ }＝0
{1}＝1
{2}＝2
{1,2}＝1

3.先求{1,2,3}的所有子集合交錯和的總和＝12
{ }＝0
{1}＝1
{2}＝2
{3}＝3
{1,2}＝1
{2,3}＝1
{1,3}＝2
{1,2,3}＝2

3.再求{1,2,3,4}的所有子集合交錯和的總和＝32
{ }＝0
{1}＝1
{2}＝2
{3}＝3
{4}=4
{1,2}＝1
{2,3}＝1
{3,4}＝1
{1,3}＝2
{2,4}＝2
{1,4}＝3
{1,2,3}＝2
{2,3,4}＝3
{1,3,4}＝2
{1,2,4}＝3
{1,2,3,4}＝2

### [問題]集合問題

ex:集合{1,2,4,6,9}的交錯和為9-6+4-2+1=6,集合{6}的交錯和為6.空集合的交錯和為0