Problem 23 - A Complete Resource Book in Mathematics for JEE Main

Question

The coefficient of \(x^{n}\) in the polynomial \(\left(x+{ }^{n} C_{0}\right)(x+3\) \(\left.{ }^{n} C_{1}\right)\left(x+5{ }^{n} C_{2}\right) \ldots\left(x+(2 n+1){ }^{n} C_{n}\right)\) is (A) \(n .2^{n}\) (B) \(n \cdot 2^{n+1}\) (C) \((n+1) \cdot 2^{n}\) (D) \(n \cdot 2^{n-1}\)

Step-by-Step Solution

Answer

The coefficient of \(x^n\) is \((n+1) \cdot 2^n\). Choice (C).

Step 1: Understanding the Problem

We are given a polynomial of the form \( \prod_{k=0}^{n} (x + (2k+1) \cdot \binom{n}{k}) \) and need to find the coefficient of \(x^n\) in the expanded polynomial.

Step 2: Set Up the Polynomial

The polynomial is expanded as \( (x + \binom{n}{0})(x + 3\binom{n}{1})(x + 5\binom{n}{2}) \ldots (x + (2n+1)\binom{n}{n}) \). This is a product of \( n+1 \) terms.

Step 3: General Expansion

The coefficient of \(x^n\) is obtained when exactly \(n\) \(x\)'s are selected from these \(n+1\) terms and one constant is chosen. This will happen when each term contributes an \(x\), except one term which contributes the constant part.

Step 4: Calculate Coefficient

For each \(k\text{-th}\) term, if the constant part \((2k+1)\binom{n}{k}\) is chosen, then from the other \(n\) terms we choose \(x\) making \(n\) terms. Thus, the coefficient is the sum of all such constants over \(k = 0\) to \(n\).

Step 5: Use Algebraic Identity

Recognize that \(\sum_{k=0}^{n} (2k+1)\binom{n}{k} = (n+1)2^{n}\) by splitting into two sums: \(2\sum_{k=0}^{n} k\binom{n}{k} + \sum_{k=0}^{n} \binom{n}{k} = (n+1)2^{n}\).

Step 6: Conclude with the Final Answer

By verifying with the algebraic identity, the coefficient of \(x^n\) is \((n+1)2^{n}\). Therefore, the correct choice is (C).

Key Concepts

Binomial Theorem

The binomial theorem is a foundational concept in algebra that provides a formula for expanding powers of binomials, which are expressions of the form \(a + b\) raised to a power \(n\). This theorem states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]where \(\binom{n}{k}\) is a binomial coefficient and can be calculated as \(\frac{n!}{k!(n-k)!}\).
The theorem simplifies the process of expanding expressions by substituting the binomial coefficients, reducing the need to manually multiply each term. It's a powerful tool for polynomial expansion, providing both insight and simplifications in algebra.
In the given problem, although not applied directly, the binomial theorem helps to understand the combination of coefficients in the expansion of the polynomial, where the selection from each term can resemble binomial multipliers in the sum representation.

Algebraic Identities

Algebraic identities are equations that hold true for all possible values of their variables. They are crucial in simplifying complicated expressions and solving problems quickly.
One powerful identity used in the solution is:\[\sum_{k=0}^{n} (2k+1)\binom{n}{k} = (n+1)2^{n}\]This identity helps to simplify the calculation of the coefficient of \(x^n\) in the given polynomial expansion by effectively summarizing complex sums into a usable form.
The identity can be split and understood through two principal parts:

• \(2\sum_{k=0}^{n} k\binom{n}{k}\)
• \(\sum_{k=0}^{n} \binom{n}{k}\)

These summations leverage combinatorial coefficients to produce a pattern recognizable in binomial expansions, making calculations much more manageable.
Using such identities allows us to bypass laborious algebraic steps, focusing instead on high-level manipulations of mathematical properties.

Combinatorics

Combinatorics deals with counting, arranging, and analyzing structures according to certain rules. It plays a pivotal role when working with polynomial expansions, especially involving binomial coefficients.
In the problem at hand, each term in the polynomial expansion involves choosing \(x\) or a constant. Combinatorics counts the different ways to choose constants from multiple expressions:- For each \(k\)-th term, the selection involves picking \(x\) terms from \(n\) and one \((2k+1)\binom{n}{k}\)- This uses a combinatorial method akin to choosing subsets or arrangements
Combinatorics also provides the formulas used to find \(\binom{n}{k}\), helping in calculating how multiple elements can be selected and arranged, optimizing results quickly in polynomial expressions.
This branch of mathematics offers a structured way to address the combination and permutation problems, crucial for understanding polynomial expansions where choice matters.