Question

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

• A. \({}^{1000}C_{501}\)
• B. \({}^{1002}C_{500}\)
• C. \({}^{1001}C_{501}\)
• D. \({}^{1002}C_{501}\)

Hint:

Whenever you see sums of shifted binomial terms, try to convert them into a single binomial using series or identities.

Answer 1

The Correct Option is D

Concept: The given expression is a sum of binomial terms. We simplify it algebraically and then use the idea that the sum of coefficients of specific powers can be obtained from a single binomial expansion.
Step 1: Rewrite the given expression The given series is: \[ \sum_{k=0}^{1000} x^k(1+x)^{1000-k} \] Factor out \((1+x)^{1000}\): \[ (1+x)^{1000}\sum_{k=0}^{1000}\left(\frac{x}{1+x}\right)^k \]
Step 2: Evaluate the geometric sum \[ \sum_{k=0}^{1000} r^k=\frac{1-r^{1001}}{1-r},\quad r=\frac{x}{1+x} \] Thus: \[ (1+x)^{1000}\cdot\frac{1-\left(\frac{x}{1+x}\right)^{1001}}{1-\frac{x}{1+x}} \] Since: \[ 1-\frac{x}{1+x}=\frac{1}{1+x} \] We get: \[ (1+x)^{1001}-x^{1001} \]
Step 3: Identify coefficients We are asked for the sum of coefficients of \(x^{499}\) and \(x^{500}\). The term \(x^{1001}\) does not affect these powers. Thus, we only consider \((1+x)^{1001}\). Coefficient of \(x^{499}\) is \({}^{1001}C_{499}\) Coefficient of \(x^{500}\) is \({}^{1001}C_{500}\)
Step 4: Add the coefficients \[ {}^{1001}C_{499}+{}^{1001}C_{500} = {}^{1002}C_{501} \] (using Pascal’s identity) Final Answer: \[ \boxed{{}^{1002}C_{501}} \]