Question

The sum of coefficients of \(x^{499}\) and \(x^{500}\) in the expression: \[ (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \cdots + x^{1000} \] is:

• A. \({}^{1001}C_{500}\)
• B. \({}^{1003}C_{501}\)
• C. \({}^{1002}C_{500}\)
• D. \({}^{1004}C_{502}\)

Hint:

Use the identity \[ {}^{n}C_{r} + {}^{n}C_{r+1} = {}^{n+1}C_{r+1} \] to quickly add adjacent binomial coefficients.

Answer 1

The Correct Option is C

Step 1: Rewrite the given expression
The given sum can be written as: \[ \sum_{k=0}^{1000} x^k (1+x)^{1000-k} \] Factor out \((1+x)^{1000}\): \[ (1+x)^{1000} \sum_{k=0}^{1000} \left(\frac{x}{1+x}\right)^k \]
Step 2: Use Geometric Series
\[ \sum_{k=0}^{1000} r^k = \frac{1-r^{1001}}{1-r} \quad \text{where } r=\frac{x}{1+x} \] Thus, \[ = (1+x)^{1000} \cdot \frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}} \] \[ = (1+x)^{1001} - x^{1001} \]
Step 3: Find coefficients of \(x^{499}\) and \(x^{500}\)
Since \(x^{1001}\) does not affect these terms, we only consider: \[ (1+x)^{1001} \] Coefficient of \(x^{499}\): \[ {}^{1001}C_{499} \] Coefficient of \(x^{500}\): \[ {}^{1001}C_{500} \]
Step 4: Add the coefficients
\[ {}^{1001}C_{499} + {}^{1001}C_{500} = {}^{1002}C_{500} \] \[ \boxed{{}^{1002}C_{500}} \]