Let `m ,n in N and g c d ( 2, n)=1.` if `30((30),(0))+((30),(1))+....+2((30),(28))+1((30),(29))=n.2^m`
then n+m is equal to
( here ` ((n),(k)) =""^(n)C_K`)

Answer

To solve the problem, we need to evaluate the expression given and find the values of \( n \) and \( m \) such that \( n + m \) can be calculated.

### Step-by-step Solution:

1. **Understanding the Expression**:
We are given the expression:
\[ S = 30 \binom{30}{0} + 29 \binom{30}{1} + 28 \binom{30}{2} + \ldots + 2 \binom{30}{28} + 1 \binom{30}{29} \]
This can be rewritten as:
\[ S = \sum_{k=0}^{29} (30-k) \binom{30}{k} \]

2. **Using the Binomial Theorem**:
The binomial theorem states that:
\[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]
We can use this to find the sum by differentiating.

3. **Differentiating the Binomial Expansion**:
Consider the expansion of \( (x + 1)^{30} \):
\[ (x + 1)^{30} = \sum_{k=0}^{30} \binom{30}{k} x^{30-k} \]
Differentiating both sides with respect to \( x \):
\[ 30(x + 1)^{29} = \sum_{k=0}^{30} (30-k) \binom{30}{k} x^{30-k-1} \]

4. **Substituting \( x = 1 \)**:
Now, substituting \( x = 1 \):
\[ 30(1 + 1)^{29} = \sum_{k=0}^{30} (30-k) \binom{30}{k} \cdot 1^{30-k-1} \]
This simplifies to:
\[ 30 \cdot 2^{29} = S \]

5. **Setting Up the Equation**:
We have found that:
\[ S = 30 \cdot 2^{29} \]
According to the problem, \( S = n \cdot 2^m \).

6. **Identifying \( n \) and \( m \)**:
From the equation \( 30 \cdot 2^{29} = n \cdot 2^m \), we can equate:
\[ n = 30 \quad \text{and} \quad m = 29 \]

7. **Checking the Condition**:
We are given that \( \gcd(2, n) = 1 \). Since \( n = 30 \) is even, we need to adjust \( n \) to satisfy this condition.

We can express \( 30 \) as:
\[ 30 = 2 \cdot 15 \]
Thus, we can rewrite \( S \) as:
\[ S = 2 \cdot 15 \cdot 2^{29} = 15 \cdot 2^{30} \]
Here, we can set:
\[ n = 15 \quad \text{and} \quad m = 30 \]

8. **Final Calculation**:
Now, we can find \( n + m \):
\[ n + m = 15 + 30 = 45 \]

### Conclusion:
The final result is:
\[ \boxed{45} \]