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To solve the problem of finding the number of ordered pairs ((A,B)) where (A\subseteq B\subseteq{1,2,\ldots,n}) and the sum of elements in (B\setminus A) is odd, we can proceed as follows:

Key Observations:

For each element (x) in ({1,\ldots,n}), there are 3 possibilities (since (A\subseteq B)):

1. (x\notin B) (case 0),
2. (x\in B\ but\ not\ in\ A) (case 1),
3. (x\in A\ and\ B) (case 2).

The sum of (B\setminus A) is the sum of elements in case 1. We need this sum to be odd.

Analysis of Even and Odd Elements:

• Even elements: Their presence in case 1 does not affect the parity of the sum (even contribution). Each even element has 3 choices (cases 0,1,2).
• Odd elements: Their presence in case 1 flips the parity of the sum. We need an odd number of odd elements in case 1.

Let (m = \floor{n/2}) (number of even elements) and (k = \ceil{n/2}) (number of odd elements).

Calculations:

• For even elements: Total choices = (3^m).
• For odd elements: The number of ways to have an odd number of elements in case 1 is (\frac{3^k -1}{2}) (using generating functions ((2+t)^k) and sum of odd coefficients).

Total ordered pairs = (3^m \times \frac{3^k -1}{2}). Since (m+k=n), this simplifies to (\frac{3^n - 3^m}{2}).

Final Formula:

[ \boxed{\frac{3^n - 3^{\floor{n/2}}}{2}} ]

Answer: (\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}}) (or simplified based on (n) parity, but this is the general form). For example, if (n) is even: (\frac{3^n -3^{n/2}}{2}); if (n) is odd: (\frac{3^n -3^{(n-1)/2}}{2}).

(\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}}) is the concise answer. However, if the problem expects an integer expression for specific (n), but since (n) is variable, this is the correct form.

Final Answer: (\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}}) (but in competition settings, it's often written as (\boxed{\frac{3^n - 3^{\frac{n - n\%2}{2}}}{2}}) or similar, but the simplest form is (\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}})).

Alternatively, for integer representation when (n) is given, but since the problem is general, the above is the answer.

(\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}})

Simplified for code or direct computation:
If (n) is even: (\frac{3^n - 3^{n/2}}{2}), else (\frac{3^n - 3^{(n-1)/2}}{2}).

(\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}}) is the correct answer.

(\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}})

Final Boxed Answer: (\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}})

But wait, in some cases, the problem might have a fixed (n), but since the user didn't specify, this is the general solution. However, if the original problem had a specific (n) (like 5), but the user input was incomplete, but based on the context, this is the solution.

Assuming the problem expects the general formula, the answer is (\boxed{\frac{3^n - 3^{\lfloor \frac{n}{2} \rfloor}}{2}}).

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