Question

If sum of Even terms are denoted by \(\mathrm{E}\) and sum of odd terms are denoted by \(\mathrm{O}\) in expansion of \((\mathrm{x}+\mathrm{a})^{\mathrm{n}}\) then \(\mathrm{O}^{2}-\mathrm{E}^{2}=\) (a) \(x^{2}-a^{2}\) (b) \(\left(\mathrm{x}^{2}-\mathrm{a}^{2}\right)^{\mathrm{n}}\) (c) \(x^{2 n}-a^{2 n}\) (b) None of these

Short answer

The short answer is \((O^2 - E^2) = (x^2 - a^2)^n\). The correct option is (b).

Step-by-step solution

Write the binomial coefficient formula

The binomial coefficient formula is given by: \[(x + a)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}a^k\]

Write the expansions for even and odd-indexed terms

In the binomial expansion, even and odd terms have alternating coefficients. We can write the sum of even terms (\(E\)) and odd terms (\(O\)) separately: \[E = \sum_{k=0}^{n/2} \binom{n}{2k}x^{n-2k}a^{2k}\] \[O = \sum_{k=0}^{(n-1)/2} \binom{n}{2k+1}x^{n-(2k+1)}a^{2k+1}\]

Express the given expression, \(O^2 - E^2\), in terms of sums

We will now express the given expression \(O^2 - E^2\) in terms of sums: \[O^2 - E^2 = \left(\sum_{k=0}^{(n-1)/2} \binom{n}{2k+1}x^{n-(2k+1)}a^{2k+1}\right)^{2} - \left(\sum_{k=0}^{n/2} \binom{n}{2k}x^{n-2k}a^{2k}\right)^{2}\]

Use the difference of squares formula

We can use the difference of squares formula, \(a^2 - b^2 = (a + b)(a - b)\), to simplify our expression: \[(O + E)(O - E) = \left(\sum_{k=0}^{(n-1)/2} \binom{n}{2k+1}x^{n-(2k+1)}a^{2k+1} + \sum_{k=0}^{n/2} \binom{n}{2k}x^{n-2k}a^{2k}\right)\left(\sum_{k=0}^{(n-1)/2} \binom{n}{2k+1}x^{n-(2k+1)}a^{2k+1} - \sum_{k=0}^{n/2} \binom{n}{2k}x^{n-2k}a^{2k}\right)\]

Factor out

We notice that both sums in the parentheses have the same power of \(x\) when summed together (or subtracted). Therefore, we can factor out the common power of \(x\) and express the difference of squares as: \[(O^2 - E^2) = (x + a)^{n}(x - a)^{n}\]

Simplify and choose the correct option

Now, we can use the fact that \((x + a)^n(x - a)^n = (x^2 - a^2)^n\) to obtain the final expression: \[O^2 - E^2 = (x^2 - a^2)^n\] This matches option (b), so the answer is: \[\boxed{(b) \,\, (x^2 - a^2)^n}\]

Key concepts

Binomial Expansion

Whenever we talk about raising a binomial expression—like \(x + a\)—to a power \(n\), we are discussing binomial expansion. This is an application of the Binomial Theorem, which assists us in expressing the binomial expression as a sum involving terms of the form \( \binom{n}{k}x^{n-k}a^k \), where \( \binom{n}{k} \) represents the binomial coefficients. The expansion of \(x + a\) raised to the power of \(n\) is written as:

\[ (x + a)^n = \sum_{k=0}^{n} \binom{n}{k}x^{n-k}a^k \]
Through this formula, one can determine each term of the expansion systematically. The practical aspect of this formula is that it aids in simplifying complex algebraic expressions, finding coefficients of particular terms, determining the total number of terms, and solving combinatorial problems.

Difference of Squares

The difference of squares is a crucial concept in algebra that refers to the factoring of expressions in the form \(a^2 - b^2\), which can be re-written as \(a + b)(a - b)\). In the context of binomial expansion, we often encounter squared terms that can be simplified using the difference of squares formula. As observed in the exercise, when we square the sums \(O\) and \(E\) and then find their difference, \(O^2 - E^2\), we can apply this formula to facilitate the simplification process:

\[ (O + E)(O - E) = O^2 - E^2 = (x + a)^n(x - a)^n \]
This formula is key for simplifying calculations, especially when dealing with polynomials, and is foundational to higher-level math such as calculus and complex numbers. In our textbook exercise, using the difference of squares rule helped in recognizing that the answer to \(O^2 - E^2\) is in fact equivalent to \(x^2 - a^2\) raised to the power of \(n\).

Binomial Coefficient

A binomial coefficient \( \binom{n}{k} \) shows up frequently in binomial expansions—it tells us the number of ways to choose \(k\) elements from a set of \(n\) distinct elements. These coefficients are also the numbers we see in Pascal's Triangle. The binomial coefficient formula is given by:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
Where \(n!\) represents \(n\) factorial, the product of all positive integers up to \(n\). For a given \(n\), the binomial coefficients form the coefficients of \(x\) and \(a\) in the terms of the binomial expansion of \(x + a\) raised to the \(n\)-th power. Understanding binomial coefficients is central to working with binomial expansions since they affect the weight of each term in the series, which is evident in how we separate out the even and odd-indexed terms to solve the given exercise.