Chapter 9: Problem 97
The coefficient of \(x^{\mathrm{n}}\) in polynomial \(\left(x+{ }^{2 \mathrm{n}+1} C_{0}\right)\left(x+{ }^{2 \mathrm{n}+1} C_{1}\right)\left(x+{ }^{2 \mathrm{n}+1} C_{2}\right) \ldots .\left(x+{ }^{2 \mathrm{n}+1} C_{\mathrm{n}}\right)\) is (A) \(2^{2 n+1}\) (B) \(2^{2 n}\) (C) \(2^{2 n-1}\) (D) none of these
Short Answer
Step by step solution
Understand the Polynomial Structure
Consider the Coefficient of x^n
Determine the Contribution of Each Choice
Calculate the Sum of All Coefficients
Find the Coefficient of x^n
Conclusion: Select the Correct Option
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Polynomial Expansion
- Identify the individual components: Powers of terms, like \(x^k\) or \(a^m\).
- Apply relevant equations: Use binomial coefficients, represented as \( \binom{n}{k} \), denoting combinations.
- Sum the terms: Add up all the derived terms to get the expanded polynomial.
Coefficient Calculation
- Choose the number of linear terms \(x\) from the available binomials; the rest will be constants.
- Use combinatorial counting methods to determine how many ways this can occur. The binomial coefficient \( \binom{n+1}{n} = n+1 \) illustrates the number of ways to choose \(n\) items from \(n+1\).
- Multiply the outcome of this combinatorial process with the sum of constant term contributions.
Combinatorics
- Combinatorics helps in counting the possible arrangements or selections of items. For instance, selecting \(n\) linear terms from \(n+1\) available factors.
- Utilizing combinations, denoted by \( \binom{n}{k} \), shows different ways to pick \(k\) items from a larger set \(n\).
- This application reduces the complexity of calculating polynomial coefficients, as seen in the coefficient calculation for \(x^n\).