Let $\{a_n\}$ be a sequence satisfying: $a_1=1$ and $a_{n+1}=2a_n+n\cdot (1+2^n),(n=1,2,3,\cdots)$. Determine the general term formula of $\{a_n\}$.
To determine the general term formula of the sequence $\{a_n\}$, we start with the given recurrence relation and initial condition:
\[ a_1 = 1 \]
\[ a_{n+1} = 2a_n + n \cdot (1 + 2^n) \quad \text{for} \quad n = 1, 2, 3, \ldots \]

1. **Define a new sequence**:
   Let's denote \( a_n = b_n + n(n-1)2^{n-2} - n - 1 \). We will find a recurrence relation for \( b_n \).

2. **Substitute \( a_n \) into the recurrence relation**:
   \[ a_{n+1} = b_{n+1} + (n+1)n2^{n-1} - (n+1) - 1 \]
   \[ 2a_n + n(1 + 2^n) = 2(b_n + n(n-1)2^{n-2} - n - 1) + n(1 + 2^n) \]

3. **Simplify the right-hand side**:
   \[ 2(b_n + n(n-1)2^{n-2} - n - 1) + n(1 + 2^n) \]
   \[ = 2b_n + 2n(n-1)2^{n-2} - 2n - 2 + n + n2^n \]
   \[ = 2b_n + n(n-1)2^{n-1} - n - 2 + n + n2^n \]
   \[ = 2b_n + n(n-1)2^{n-1} - 2 + n2^n \]

4. **Equate the expressions**:
   \[ b_{n+1} + (n+1)n2^{n-1} - (n+1) - 1 = 2b_n + n(n-1)2^{n-1} - 2 + n2^n \]

5. **Isolate \( b_{n+1} \)**:
   \[ b_{n+1} = 2b_n \]

6. **Solve the recurrence relation for \( b_n \)**:
   Given \( b_1 = 3 \), we have:
   \[ b_n = 3 \cdot 2^{n-1} \]

7. **Substitute \( b_n \) back into the expression for \( a_n \)**:
   \[ a_n = b_n + n(n-1)2^{n-2} - n - 1 \]
   \[ a_n = 3 \cdot 2^{n-1} + n(n-1)2^{n-2} - n - 1 \]

8. **Simplify the expression**:
   \[ a_n = 3 \cdot 2^{n-1} + \frac{n(n-1)}{2} \cdot 2^{n-1} - n - 1 \]
   \[ a_n = 3 \cdot 2^{n-1} + \frac{n(n-1)}{2} \cdot 2^{n-1} - n - 1 \]
   \[ a_n = 2^{n-1} \left( 3 + \frac{n(n-1)}{2} \right) - n - 1 \]
   \[ a_n = 2^{n-1} \left( 3 + \frac{n^2 - n}{2} \right) - n - 1 \]
   \[ a_n = 2^{n-1} \left( \frac{6 + n^2 - n}{2} \right) - n - 1 \]
   \[ a_n = 2^{n-1} \left( \frac{n^2 - n + 6}{2} \right) - n - 1 \]
   \[ a_n = 2^{n-1} \cdot \frac{n^2 - n + 6}{2} - n - 1 \]
   \[ a_n = 2^{n-2} (n^2 - n + 6) - n - 1 \]

The final answer is \(\boxed{a_n = 2^{n-2} (n^2 - n + 6) - n - 1}\).