What is `""^(n)C_(1)+ ""^(n)C_(2)+……… + ""^(n)C_(n)` ?

Answer

To solve the problem, we need to find the sum of the binomial coefficients from \( nC_1 \) to \( nC_n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficient Identity**:
We start with the well-known identity for binomial coefficients:
\[ \sum_{k=0}^{n} {n \choose k} = 2^n \]
This identity states that the sum of all binomial coefficients for a given \( n \) is equal to \( 2^n \).

2. **Breaking Down the Sum**:
We can express the sum of the binomial coefficients from \( nC_1 \) to \( nC_n \) as:
\[ {n \choose 1} + {n \choose 2} + \ldots + {n \choose n} = \sum_{k=1}^{n} {n \choose k} \]
We notice that this sum excludes the term \( {n \choose 0} \).

3. **Using the Identity**:
From the identity we stated earlier, we can rewrite the sum as:
\[ \sum_{k=1}^{n} {n \choose k} = \sum_{k=0}^{n} {n \choose k} - {n \choose 0} \]
Here, \( {n \choose 0} = 1 \).

4. **Substituting the Values**:
Now, substituting the known values into the equation:
\[ \sum_{k=1}^{n} {n \choose k} = 2^n - 1 \]

5. **Final Result**:
Therefore, the sum \( {n \choose 1} + {n \choose 2} + \ldots + {n \choose n} \) simplifies to:
\[ 2^n - 1 \]

### Conclusion:
The final answer is:
\[ {n \choose 1} + {n \choose 2} + \ldots + {n \choose n} = 2^n - 1 \]