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

To determine the sum of the coefficients of \(x^{499}\) and \(x^{500}\) in the given expression, let's analyze the expansion in the problem.

The series given is:

\[ (1+x)^{1000} + x(1+x)^{999} + x^2(1+x)^{998} + \cdots + x^{1000} \]

This is a form of a generating function with each term being weighted by a power of \(x\):

The first term: \((1+x)^{1000}\) contributes coefficients according to the binomial expansion:

\((1+x)^{1000} = \sum_{n=0}^{1000} \binom{1000}{n} x^n\)

The second term: \(x(1+x)^{999}\) is effectively a shift by 1 power of \(x\), so its expansion becomes:

\((x)(1+x)^{999} = x \sum_{n=0}^{999} \binom{999}{n} x^n = \sum_{n=0}^{999} \binom{999}{n} x^{n+1}\)

Continuing this pattern, the \(k\)-th term is:

\(x^k(1+x)^{1000-k} = x^k \sum_{n=0}^{1000-k} \binom{1000-k}{n} x^n = \sum_{n=0}^{1000-k} \binom{1000-k}{n} x^{n+k}\)

Thus, the coefficient of \(x^m\) in the full expansion is given by adding the shifted binomial terms:

\(\sum_{k=0}^{1000} \binom{1000-k}{m-k}\) where \(0 \leq k \leq m\).

Now, let's find the coefficient of \(x^{499}\) and \(x^{500}\) by focusing on possible values of \(k\):

• For \(x^{499}\):
  Since \(0 \leq k \leq 499\), the coefficient is: \[ \sum_{k=0}^{499} \binom{1000-k}{499-k} \] which is: \[ \sum_{r=0}^{499} \binom{1001-r}{500} \] This uses the property of shifted index \(m-k=r\), thus rewriting in sum directly to the formula.
• For \(x^{500}\):
  Similarly, for \(x^{500}\): \[ \sum_{k=0}^{500} \binom{1000-k}{500-k} \] which converts: \[ \sum_{r=0}^{500} \binom{1001-r}{501} \]

The sum of these coefficients converts through the series binomial identity and simplifications (e.g., Vandermonde's identity) as:

\[ \sum_{k=0}^{1000} \binom{1000-k}{999-k} = \binom{1002}{500} \]

Finally, verifying options, we find that the result matches with \({}^{1002}C_{500}\). Therefore, the sum of the coefficients of \(x^{499}\) and \(x^{500}\) is \(\boxed{^{1002}C_{500}}\).