Question

The coefficient of \( x^{50} \) in the expression of \[ (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + \ldots + 1001x^{1000} \]

• A. \( \binom{1005}{50} \)
• B. \( \binom{1005}{48} \)
• C. \( \binom{1002}{50} \)
• D. \( \binom{1002}{51} \)

Hint:

For binomial expansions in a series, match the degree of the terms to find the relevant coefficient.

Answer 1

The Correct Option is C

Step 1: The given series is a sum of terms of the form: \[ kx^{1001-k}(1 + x)^{1000-k} \] where \( k = 1, 2, 3, \ldots, 1001 \). Step 2: We need to find the coefficient of \( x^{50} \) in this series. We first focus on the term \( (1 + x)^{1000-k} \). Expanding this term using the binomial expansion gives: \[ (1 + x)^{1000-k} = \sum_{r=0}^{1000-k} \binom{1000-k}{r} x^r \] So the coefficient of \( x^{50} \) in this expansion is \( \binom{1000-k}{50} \) when \( 1000-k \geq 50 \). Step 3: Now consider the term \( kx^{1001-k}(1 + x)^{1000-k} \). The degree of \( x \) in the product is \( 1001-k + r \). We are looking for \( 1001 - k + r = 50 \), so: \[ r = 50 - (1001 - k) = k - 951 \] Step 4: To ensure the term is valid, we require \( 0 \leq k - 951 \leq 1000-k \). Solving this condition, we find that \( k = 50 \). Thus, the coefficient of \( x^{50} \) is \( \binom{1002}{50} \). Thus, the correct answer is (c) \( \binom{1002}{50} \).