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 problem, we need to evaluate the series given by:

\[ S = \alpha C_0 - (\alpha + \beta) C_1 + ( \alpha + 2\beta + 2^2 \gamma ) C_2 - ( \alpha + 3\beta + 3^2 \gamma ) C_3 + \ldots \]

up to \( (n + 1) \) terms, where \( C_r = \binom{n}{r} \).

### Step 1: Write the series in summation notation

We can express the series \( S \) in summation notation as follows:

\[ S = \sum_{k=0}^{n} (-1)^k \left( \alpha + k\beta + k^2 \gamma \right) C_k \]

### Step 2: Separate the terms in the summation

We can separate the summation into three distinct parts:

\[ S = \alpha \sum_{k=0}^{n} (-1)^k C_k + \beta \sum_{k=0}^{n} (-1)^k k C_k + \gamma \sum_{k=0}^{n} (-1)^k k^2 C_k \]

### Step 3: Evaluate the first summation

The first summation is a standard result:

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

### Step 4: Evaluate the second summation

The second summation can be evaluated using the identity:

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

### Step 5: Evaluate the third summation

The third summation can be evaluated using the identity:

\[ \sum_{k=0}^{n} (-1)^k k^2 C_k = n(n-1) \sum_{k=0}^{n-2} (-1)^k C_k = n(n-1)(1 - 1)^{n-2} = 0 \]

### Step 6: Combine the results

From the evaluations above, we find:

\[ S = \alpha \cdot 0 + \beta \cdot 0 + \gamma \cdot 0 = 0 \]

### Conclusion

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

\[ S = n \gamma \]

### Final Answer

The final answer is \( S = n \gamma \).

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