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 given expression \( \Sigma_{i=1}^{n} \left( {n+1 \choose i} - {n \choose i} \right) \), we can break it down step by step.

### Step 1: Rewrite the Summation
We start with the expression:
\[ \Sigma_{i=1}^{n} \left( {n+1 \choose i} - {n \choose i} \right) \]
This can be separated into two summations:
\[ \Sigma_{i=1}^{n} {n+1 \choose i} - \Sigma_{i=1}^{n} {n \choose i} \]

### Step 2: Evaluate Each Summation
The second summation \( \Sigma_{i=1}^{n} {n \choose i} \) can be evaluated using the binomial theorem:
\[ \Sigma_{i=0}^{n} {n \choose i} = 2^n \]
Thus,
\[ \Sigma_{i=1}^{n} {n \choose i} = 2^n - {n \choose 0} = 2^n - 1 \]

### Step 3: Evaluate the First Summation
For the first summation \( \Sigma_{i=1}^{n} {n+1 \choose i} \), we apply the same logic:
\[ \Sigma_{i=0}^{n+1} {n+1 \choose i} = 2^{n+1} \]
Thus,
\[ \Sigma_{i=1}^{n} {n+1 \choose i} = 2^{n+1} - {n+1 \choose 0} = 2^{n+1} - 1 \]

### Step 4: Combine the Results
Now we can substitute back into our original expression:
\[ \Sigma_{i=1}^{n} \left( {n+1 \choose i} - {n \choose i} \right) = (2^{n+1} - 1) - (2^n - 1) \]
This simplifies to:
\[ 2^{n+1} - 1 - 2^n + 1 = 2^{n+1} - 2^n \]

### Step 5: Factor the Expression
We can factor out \( 2^n \):
\[ 2^{n+1} - 2^n = 2^n(2 - 1) = 2^n \]

### Final Answer
Thus, the value of the given expression is:
\[ \boxed{2^n} \]