The coefficient of $x^{301}$ in $(1+x)^{500}+x(1+x)^{499}+x^2(1+x)^{498}+\ldots \ldots +x^{500}$ is :
- ${ }^{500} C _{301}$
- ${ }^{501} C_{200}$
- ${ }_3{ }^{500} C_{300}$
- ${ }^{501} C_{302}$
The Correct Option is B
Approach Solution - 1
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 \({}^{501}C_{200}\).
Approach Solution -2
Coefficient of in is given by
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Concepts Used:
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