Question

The coefficient of $x^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots \ldots +x^{500}$ is :

• A. ${ }^{500} C _{301}$
• B. ${ }^{501} C_{200}$
• C. ${ }_3{ }^{500} C_{300}$
• D. ${ }^{501} C_{302}$

Answer 1

The Correct Option is B

To find the coefficient of \(x^{301}\) in the given expression, let's first analyze the series:

The series is:

\((1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \ldots + x^{500}\)

This can be rewritten as:

\(\sum_{k=0}^{500} x^k (1+x)^{500-k}\)

The general term of this series is:

\(x^k \cdot (1+x)^{500-k}\)

In this term, to get the coefficient of \(x^{301}\), we need:

\(k + r = 301\)

where \(r\) is the power of \(x\) from the expansion of \((1+x)^{500-k}\).

Thus, we find \(r\) from the expansion of \((1+x)^{500-k}\) as:

\(\binom{500-k}{r}\)

Therefore, we need:

\(k + r = 301\)

Moreover, we must have \(r + k = 301\), so \(r = 301 - k\).

Thus, the coefficient of \(x^{301}\) in the general term is:

\(\binom{500-k}{301-k}\)

Sum over all \(k\) from 0 to 500 gives:

\(\sum_{k=0}^{500} \binom{500-k}{301-k} \)

Switch variables \(j = 500 - k\):

So, the sum becomes:

\(\sum_{j=0}^{500} \binom{j}{301-(500-j)} = \sum_{j=0}^{500} \binom{j}{301-j} \)

Further simplifying using the binomial coefficient identity:

\(\sum_{j=0}^{500} \binom{j}{200} = \binom{501}{200} \)

Thus, the coefficient of \(x^{301}\) is: \(\mathbf{\binom{501}{200}}\)

This matches with the answer option provided:

${ }^{501} C_{200}$