The value of `(n+2).^(n)C_(0).2^(n+1)-(n+1).^(n)C_(1).2^(n)+(n).^(n)C_(2).2^(n-1)-….." to " (n+1)` terms is equal to

Options

A. 4
B. 4n
C. `4(n+1)`
D. `2(n+2)`

Answer

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

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

This continues for \( (n+1) \) terms.

### Step 1: Identify the General Term
The general term of the series can be expressed as:

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

where \( r \) ranges from \( 0 \) to \( n+1 \).

### Step 2: Write the Series
The series can be rewritten as:

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

### Step 3: Split the Sum
We can split the sum into two parts:

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

### Step 4: Evaluate the First Sum
The first sum can be simplified using the binomial theorem:

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

### Step 5: Evaluate the Second Sum
The second sum can be evaluated using the identity \( r \cdot \binom{n}{r} = n \cdot \binom{n-1}{r-1} \):

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

This simplifies to:

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

### Step 6: Combine the Results
Combining both results, we have:

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

### Step 7: Simplify the Expression
This can be simplified to:

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

### Step 8: Final Result
Thus, the value of the expression is:

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

### Conclusion
The value of the given expression is \( 2(n + 1)(-1)^{n} \).