If `C_(0), C_(1), C_(2), …. C_(n)` denote the coefficients in the expansion of `(1+x)^(n)`, then the value of `sum_(r=0)^(n) (r+1)C_(r )` is

Options

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

Answer

To solve the problem, we need to find the value of the sum:

\[ \sum_{r=0}^{n} (r+1) C_r \]

where \( C_r \) are the coefficients in the expansion of \( (1+x)^n \).

### Step 1: Understanding the Coefficients
The coefficients \( C_r \) correspond to the binomial coefficients, which can be expressed as:

\[ C_r = \binom{n}{r} \]

Thus, we can rewrite the sum as:

\[ \sum_{r=0}^{n} (r+1) \binom{n}{r} \]

### Step 2: Splitting the Sum
We can split the sum into two parts:

\[ \sum_{r=0}^{n} (r+1) \binom{n}{r} = \sum_{r=0}^{n} r \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} \]

### Step 3: Evaluating the Second Sum
The second sum is straightforward:

\[ \sum_{r=0}^{n} \binom{n}{r} = 2^n \]

This is a well-known result from the binomial theorem.

### Step 4: Evaluating the First Sum
For the first sum \( \sum_{r=0}^{n} r \binom{n}{r} \), we can use the identity:

\[ r \binom{n}{r} = n \binom{n-1}{r-1} \]

Thus, we can rewrite the sum as:

\[ \sum_{r=0}^{n} r \binom{n}{r} = n \sum_{r=1}^{n} \binom{n-1}{r-1} = n \sum_{k=0}^{n-1} \binom{n-1}{k} = n \cdot 2^{n-1} \]

### Step 5: Combining the Results
Now we can combine both parts of the sum:

\[ \sum_{r=0}^{n} (r+1) \binom{n}{r} = n \cdot 2^{n-1} + 2^n \]

### Step 6: Simplifying the Expression
We can factor out \( 2^{n-1} \):

\[ = 2^{n-1} (n + 2) \]

### Final Result
Thus, the value of the sum is:

\[ \sum_{r=0}^{n} (r+1) C_r = (n + 2) \cdot 2^{n-1} \]