The value of `Sigma_(i=1)^(n)(.^(n+1)C_(i)-.^(n)C_(i))` is equal to

Options

A. `2^(n)`
B. `2^(n)+1`
C. `3.2^(n)`
D. `2^(n)-1`

Answer

To solve the problem, we need to evaluate the summation:

\[ S_n = \sum_{i=1}^{n} \left( \binom{n+1}{i} - \binom{n}{i} \right) \]

### Step-by-Step Solution:

1. **Rewrite the Summation**:
We can break down the summation into two parts:
\[ S_n = \sum_{i=1}^{n} \binom{n+1}{i} - \sum_{i=1}^{n} \binom{n}{i} \]

2. **Evaluate Each Summation**:
- The first summation, \(\sum_{i=1}^{n} \binom{n+1}{i}\), can be evaluated using the binomial theorem. According to the binomial theorem:
\[ (1 + x)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} x^k \]
Setting \(x = 1\):
\[ 2^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} \]
Therefore:
\[ \sum_{i=1}^{n} \binom{n+1}{i} = 2^{n+1} - \binom{n+1}{0} - \binom{n+1}{n+1} = 2^{n+1} - 1 - 1 = 2^{n+1} - 2 \]

- The second summation, \(\sum_{i=1}^{n} \binom{n}{i}\), can also be evaluated similarly:
\[ (1 + x)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^k \]
Setting \(x = 1\):
\[ 2^{n} = \sum_{k=0}^{n} \binom{n}{k} \]
Therefore:
\[ \sum_{i=1}^{n} \binom{n}{i} = 2^{n} - \binom{n}{0} = 2^{n} - 1 \]

3. **Combine the Results**:
Now we can substitute back into our expression for \(S_n\):
\[ S_n = (2^{n+1} - 2) - (2^{n} - 1) \]
Simplifying this gives:
\[ S_n = 2^{n+1} - 2 - 2^{n} + 1 = 2^{n+1} - 2^{n} - 1 \]
Factoring out \(2^{n}\):
\[ S_n = 2^{n}(2 - 1) - 1 = 2^{n} - 1 \]

4. **Final Result**:
Thus, the value of the summation is:
\[ S_n = 2^{n} - 1 \]