Problem 5 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Sum of odd terms is \(A\) and sum of even terms is \(B\) in the expansion \((x+a)^{n}\), then [RPET - 1987, 1992; UPSEAT - 2004; Roorkee - 1986] (a) \(A B=\frac{1}{4}\left[(x-a)^{2 n}-(x+a)^{2 n}\right]\) (b) \(2 A B=(x+a)^{2 n}-(x-a)^{2 n}\)(c) \(4 A B=(x+a)^{2 n}-(x-a)^{2 n}\) (d) none of these

Step-by-Step Solution

Answer

The correct answer is (a).

Step 1: Understand the Binomial Expansion

The binomial expansion of (x+a)^n is expressed as \((x+a)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k\). Here, \(\binom{n}{k}\) represents the binomial coefficient.

Step 2: Identify Odd and Even Terms

In the expansion, terms with odd values of \(k\) will contribute to the odd sum \(A\), and terms with even values \(k\) will contribute to the even sum \(B\). The odd terms are \(\sum_{k \text{ is odd}} \binom{n}{k} x^{n-k} a^k\) and even terms are \(\sum_{k \text{ is even}} \binom{n}{k} x^{n-k} a^k\).

Step 3: Express Sum of Odd Terms (A)

For odd \(k\), \(A\) can be calculated using: \[ A = \frac{(x+a)^n - (x-a)^n}{2} \]This formula comes from the fact that \((x+a)^n\) includes all terms and \((x-a)^n\) has terms with alternating signs.

Step 4: Express Sum of Even Terms (B)

For even \(k\), \(B\) can be calculated using: \[ B = \frac{(x+a)^n + (x-a)^n}{2} \]This formula is derived because even terms retain the same sign in both expansions \((x+a)^n\) and \((x-a)^n\).

Step 5: Calculate the Product AB

Substituting \(A\) and \(B\) we found:\[ A \cdot B = \left( \frac{(x+a)^n - (x-a)^n}{2} \right) \cdot \left( \frac{(x+a)^n + (x-a)^n}{2} \right) = \frac{1}{4} \left( (x+a)^{2n} - (x-a)^{2n} \right) \] Using the difference of squares formula: \( (p-q)(p+q) = p^2 - q^2 \).

Step 6: Verify the Correct Choice

From the calculated expression, we identify that the correct option is (a): \[ AB = \frac{1}{4} \left( (x-a)^{2n} - (x+a)^{2n} \right) \]

Key Concepts

Sum of Odd and Even Terms

In the expansion of a binomial expression, such as \((x+a)^n\), each term contributes differently based on whether it arises from an odd or even exponent. When examining the sum of odd and even terms, it's crucial to differentiate between the two.

- **Odd Terms:** These are terms in the expansion where the exponent of one of the variables (either \(x\) or \(a\)) is odd. In our case, it pertains to the terms where the exponent \(k\) in \(a^k\) is odd.- **Even Terms:** These consist of terms where the exponent \(k\) in \(a^k\) is even.

In order to determine the contribution of either type, we can use symmetry principles. The expression for the sum of odd terms \(A\) is derived by removing the component of \((x-a)^n\) from \((x+a)^n\). This works because \((x-a)^n\), being a mirror image with alternating signs, cancels out the even terms. The sum \(A\) is thus expressed by:\[ A = \frac{(x+a)^n - (x-a)^n}{2} \]

For the even terms sum \(B\), both \((x+a)^n\) and \((x-a)^n\) contribute terms in the series with the same sign, thus enhancing their presence in the expansion:\[ B = \frac{(x+a)^n + (x-a)^n}{2} \]

Binomial Coefficient

The binomial coefficient, denoted as \(\binom{n}{k}\), plays a pivotal role in binomial expansions. It is a value that represents the number of ways to choose \(k\) elements out of \(n\), without considering the order. The formula to calculate it is given by:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Here, \(n!\) signifies factorial, the product of all positive integers up to \(n\).

In essence, these coefficients determine the weight or magnitude of each term in the expansion \((x+a)^n\). For example, for \((x+a)^3\), the expansion is:\[ (x+a)^3 = \binom{3}{0}x^3 + \binom{3}{1}x^2a + \binom{3}{2}xa^2 + \binom{3}{3}a^3 \]

In this expression:

• \(\binom{3}{0} = 1\)
• \(\binom{3}{1} = 3\)
• \(\binom{3}{2} = 3\)
• \(\binom{3}{3} = 1\)

These coefficients indicate the frequency of each associated term throughout the expansion, thereby providing its structure and ensuring accuracy in calculations.

Difference of Squares Formula

The difference of squares formula is a fundamental algebraic identity that describes the product of a sum and a difference of two terms.

This identity is expressed as:\[ (p+q)(p-q) = p^2 - q^2 \]

The formula is invaluable for simplifying expressions, especially when dealing with binomials of the form \((x+a)^n - (x-a)^n\). By viewing this subtraction through the lens of the difference of squares, we can set \(p = (x+a)^n\) and \(q = (x-a)^n\), which yields:\[ (x+a)^n(x-a)^n - (x-a)^n(x+a)^n = ((x+a)^n - (x-a)^n)((x+a)^n + (x-a)^n) = (x+a)^{2n} - (x-a)^{2n} \]

This elegant identity simplifies otherwise complex polynomial manipulations and is a powerful tool in solving advanced algebraic problems. It highlights the connection between algebraic structures and the deep, underlying symmetry in mathematical systems.