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

We are asked to find the coefficient of \(x^{301}\) in the expansion of the series:

\[ (1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \cdots + x^{500}. \]

Step 1: Understanding the Series

The given series can be written as:

\[ S = (1+x)^{500} + x(1+x)^{499} + x^2(1+x)^{498} + \cdots + x^{500}. \]

This is a sum of terms where each term involves \((1+x)^{500-n}\) multiplied by \(x^n\), where \(n\) ranges from 0 to 500.

The general term in the expansion is:

\[ x^n(1+x)^{500-n}. \]

Step 2: Finding the Coefficient of \(x^{301}\)

We need to find the coefficient of \(x^{301}\) in the entire series. For each term \(x^n(1+x)^{500-n}\), the exponent of \(x\) in the expanded form of \((1+x)^{500-n}\) will be \(500-n\). We are interested in terms where the total exponent of \(x\) equals 301.

The exponent of \(x\) in each term is:

\[ n + k = 301, \]

where \(k\) is the exponent of \(x\) in the expansion of \((1+x)^{500-n}\). The coefficient of \(x^{301}\) in \((1+x)^{500-n}\) is:

\[ {}^{500-n}C_{301-n}. \]

Step 3: Coefficient of \(x^{301}\) in the Series

Thus, the required coefficient is the sum of the coefficients of \(x^{301}\) in all terms. After simplifying, we get the coefficient of \(x^{301}\) as:

\[ {}^{501}C_{200}. \]

Thus, the correct answer is \({}^{501}C_{200}\).

Answer 2

$(1+x{)}^{500}+x(1+x{)}^{499}+x^2(1+x{)}^{498}+\dots +x^{500}$
$=(1+x{)}^{500}\cdot \left\{\frac{1-{\left(\frac{x}{1+x}\right)}^{501}}{1-\frac{x}{1+x}}\right\}$
$=(1+x{)}^{500}\frac{\left((1+x{)}^{501}-x^{501}\right)}{(1+x{)}^{501}}\cdot (1+x)$
$=(1+x{)}^{501}-x^{501}$
Coefficient of $x^{301}$ in $(1+x{)}^{501}-x^{501}$ is given by
${}^{501}{C}_{301}={}^{501}{C}_{200}$