Problem 81 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \((1+x)^{\mathrm{m}}=C_{0}+C_{1} x+C_{2} x^{2}+\ldots+C_{\mathrm{n}} x^{\mathrm{n}}\), then for \(n\) even, \(C_{0}^{2}-C_{1}^{2}+C_{2}^{2}-\ldots+(-1)^{n} C_{n}^{2}\) is equal to (A) 0 (B) \((-1)^{n / 2 \cdot n} C_{n / 2}\) (C) \({ }^{n} C_{n / 2}\) (D) none of these

Step-by-Step Solution

Answer

(B) \((-1)^{n/2} \cdot C_{n/2}\) applies when \(n\) is even.

Step 1: Interpret the Given Expression

The expression \((1+x)^m\) is expanded as \(C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n\), where each \(C_k\) is a binomial coefficient. This indicates that \(C_k = \binom{m}{k}\).

Step 2: Evaluate Sum Expression

We need to find the value of \(C_0^2 - C_1^2 + C_2^2 - \ldots + (-1)^n C_n^2\). This series involves squared coefficients of the binomial expansion.

Step 3: Use Generating Functions

To solve \(C_0^2 - C_1^2 + C_2^2 - \ldots + (-1)^n C_n^2\), consider the generating function \((1+x)^m (1-x)^m = (1-x^2)^m\). This will give us the sum of alternate squared coefficients.

Step 4: Simplification Using Binomial Theorem

Because \((1-x^2)^m\) expands as \( \sum_{k=0}^{m} (-1)^k \binom{m}{k} x^{2k}\), compare it with the provided expression. It demonstrates that the sum \(C_0^2 - C_1^2 + C_2^2 - \ldots + (-1)^n C_n^2\) equals \((-1)^{n/2} \cdot \binom{m}{n/2}\) for \(n\) even.

Step 5: Conclusion with Choice Verification

The solution matches option B in the given multiple-choice problem. Thus, the final expression simplifies to \((-1)^{n/2} \cdot n\) times the middle binomial coefficient when \(n\) is even.

Key Concepts

Binomial Expansion

The Binomial Expansion is a way of expressing powers of a binomial, which is an algebraic expression that contains two terms added or subtracted together. For example, if you have the expression \((1+x)^m\), it represents a binomial raised to the power of \(m\). This expansion can be written as a series of terms:

• \(C_0\)
• \(C_1 x\)
• \(C_2 x^2\)
• ... \(C_n x^n\)

Each term in the expansion has a coefficient, which is called a Binomial Coefficient. The general term in a binomial expansion is given by \(C_k x^k\) where \(k\) is the current term number in the sequence. Understanding how to perform binomial expansion can simplify equations and help in deriving more complex mathematical concepts.

Binomial Coefficients

The coefficients you find in the expression of a binomial expansion are known as Binomial Coefficients. They are typically denoted as \(C_k\) or \(\binom{m}{k}\), where \(m\) is the power to which the binomial is raised, and \(k\) represents the specific term number. These coefficients show the number of ways \(k\) items can be chosen from \(m\) items, and they follow the pattern found in Pascal's Triangle.
When working with problems that involve binomial coefficients, always remember:

• \(\binom{m}{0} = 1\)
• \(\binom{m}{1} = m\)
• \(\binom{m}{k} = \frac{m!}{k!(m-k)!}\)

These coefficients become important when you need to find specific terms of a binomial expansion or when calculating probabilities in combinatorial situations.

Generating Functions

Generating Functions are a powerful tool used to study sequences and can transform a sequence of numbers into a function. For example, in the provided solution, the generating function \((1+x)^m (1-x)^m = (1-x^2)^m\) was used to solve for the alternate squared coefficients of the binomial expansion.
This function provides a new way to look at problems by allowing you to sum up an infinite series or understand the structure of coefficients in a polynomial expansion.

• They help in solving recurrence relations
• Reconciling patterns in sequence
• Understanding combinatorial identities

In this exercise, generating functions were used to find the sum of alternate squared coefficients, ultimately leading to the correct answer. Practicing with generating functions is a great way to deepen your understanding of both simple and complex sequences.