If `n gt2` and `alpha, beta, gamma in R`, then the value of `S=alphaC_(0)-(alpha+beta)C_(1)+(alpha+2beta+2^(2)gamma)C_(2)-(alpha+3beta+3^(2)gamma)C_(3)+…." upto "(n+1)` terms is equal to
(where, `C_(r)` denotes `.^(n)C_(r)`)

Options

A. 0
B. `2^(n-2)gamma`
C. `n^(2)2^(n-2)gamma`
D. ngamma`

Answer

To solve the given problem, we need to evaluate the sum

\[ S = \sum_{k=0}^{n} (-1)^k P_k C_k \]

where \( P_k \) is defined as:

- \( P_0 = \alpha \)
- \( P_1 = -(\alpha + \beta) \)
- \( P_2 = \alpha + 2\beta + 2^2\gamma \)
- \( P_3 = -(\alpha + 3\beta + 3^2\gamma) \)
- and so on, up to \( n+1 \) terms.

### Step 1: Identify the structure of the sum

The sum alternates in sign and involves binomial coefficients \( C_k \). We can express the sum \( S \) as:

\[ S = \sum_{k=0}^{n} (-1)^k P_k C_k \]

### Step 2: Break down \( P_k \)

From the problem, we see that \( P_k \) can be broken down into three parts based on the coefficients of \( \alpha \), \( \beta \), and \( \gamma \):

1. The coefficient of \( \alpha \) contributes \( \alpha C_0 \).
2. The coefficient of \( \beta \) contributes \( -(\alpha + \beta) C_1 + (2\beta + 2^2\gamma) C_2 - (3\beta + 3^2\gamma) C_3 + \ldots \).
3. The coefficient of \( \gamma \) contributes \( (2^2\gamma) C_2 - (3^2\gamma) C_3 + \ldots \).

### Step 3: Use known summation results

We can use the known results for sums involving binomial coefficients. Specifically, we know:

1. \(\sum_{k=0}^{n} (-1)^k C_k = 0\)
2. \(\sum_{k=1}^{n} (-1)^k k C_k = 0\)
3. \(\sum_{k=2}^{n} (-1)^k k(k-1) C_k = 0\)

### Step 4: Evaluate the contributions

From the above results, we can evaluate the contributions:

- The contribution from \( \alpha \) is \( \alpha \cdot 0 = 0 \).
- The contribution from \( \beta \) is \( \beta \cdot 0 = 0 \).
- The contribution from \( \gamma \) is also \( 0 \).

### Step 5: Identify the remaining term

The only term that remains is from \( C_1 \):

\[ S = \sum_{k=1}^{n} (-1)^k C_k \]

This simplifies to \( n \) because the only non-zero contribution comes from \( C_1 \).

### Final Result

Thus, the value of \( S \) is:

\[ S = n \]