The value of the sum `""^(1000)C_(50) + ""^(999)C_(49) +""^(998)C_(48)+"…."""^(950)C_(0)` is

Options

A. (a) `""^(1001)C_(50)`
B. (b) `""^(1002)C_(951)`
C. (c) `""^(1001)C_(950)`
D. (d) `""^(1002)C_(50)`

Explanation

To find the value of the sum \( \binom{1000}{50} + \binom{999}{49} + \binom{998}{48} + \ldots + \binom{950}{0} \), we can use the properties of binomial coefficients and the Binomial Theorem.

### Step-by-step Solution:

1. **Understanding the Sum**: The sum we need to evaluate is:
\[ S = \binom{1000}{50} + \binom{999}{49} + \binom{998}{48} + \ldots + \binom{950}{0} \]
This sum consists of binomial coefficients where the upper index decreases from 1000 to 950 and the lower index decreases from 50 to 0.

2. **Using the Binomial Theorem**: We know from the Binomial Theorem that:
\[ (1 + x)^n = \sum_{k=0}^{n} \binom{n}{k} x^k \]
We can express our sum in terms of the coefficients of a polynomial.

3. **Finding the Coefficient**: The sum \( S \) can be interpreted as the coefficient of \( x^{950} \) in the expansion of:
\[ (1 + x)^{1000} + (1 + x)^{999} + (1 + x)^{998} + \ldots + (1 + x)^{950} \]
This is because each term \( \binom{n}{k} \) corresponds to the coefficient of \( x^k \) in \( (1 + x)^n \).

4. **Simplifying the Expression**: The above expression can be simplified using the formula for the sum of a geometric series. The sum can be expressed as:
\[ \sum_{n=950}^{1000} (1 + x)^n = (1 + x)^{950} \left(1 + (1 + x) + (1 + x)^2 + \ldots + (1 + x)^{50}\right) \]
The series inside the parentheses is a geometric series with first term 1 and common ratio \( (1 + x) \), and it has 51 terms.

5. **Calculating the Geometric Series**: The sum of the geometric series is given by:
\[ \frac{(1 + x)^{51} - 1}{(1 + x) - 1} = \frac{(1 + x)^{51} - 1}{x} \]

6. **Final Expression**: Therefore, we can write:
\[ S = (1 + x)^{950} \cdot \frac{(1 + x)^{51} - 1}{x} \]

7. **Finding the Coefficient of \( x^{950} \)**: We need to find the coefficient of \( x^{950} \) in:
\[ (1 + x)^{1001} - (1 + x)^{950} \]
The coefficient of \( x^{950} \) in \( (1 + x)^{1001} \) is \( \binom{1001}{950} = \binom{1001}{51} \).

8. **Final Result**: Thus, the value of the sum \( S \) is:
\[ S = \binom{1001}{51} \]

### Conclusion:
The value of the sum \( \binom{1000}{50} + \binom{999}{49} + \ldots + \binom{950}{0} \) is \( \binom{1001}{51} \).