Understanding Binomial Expansion and Number of Terms

The question asks for the number of terms in the expansion of a given algebraic expression. To find the number of terms, we first need to simplify the expression into a standard form, preferably a binomial raised to a power.

The given expression is: \( \left\{(2x – 3y)^2 (2x + 3y)^2\right\}^2 \)

Step-by-Step Simplification of the Expression

Let's simplify the expression step by step:

Focus on the inner part of the expression: \( (2x – 3y)^2 (2x + 3y)^2 \)

We can use the property of exponents which states that \(a^m b^m = (ab)^m\). Applying this, we get: \( (2x – 3y)^2 (2x + 3y)^2 = \left[(2x – 3y)(2x + 3y)\right]^2 \)

Now, consider the term inside the square brackets: \( (2x – 3y)(2x + 3y) \). This is in the form of \((a-b)(a+b)\), which expands to \(a^2 - b^2\).

Here, \(a = 2x\) and \(b = 3y\). So, \( (2x – 3y)(2x + 3y) = (2x)^2 - (3y)^2 \)

Calculating the squares: \( (2x)^2 = 4x^2 \) and \( (3y)^2 = 9y^2 \).

Substituting these back, we get: \( (2x – 3y)(2x + 3y) = 4x^2 - 9y^2 \)

Now substitute this back into the expression from step 2: \( \left[4x^2 - 9y^2\right]^2 \)

The original expression was \( \left\{(2x – 3y)^2 (2x + 3y)^2\right\}^2 \). Substituting the simplified inner part, we get: \( \left[ (4x^2 - 9y^2)^2 \right]^2 \)

Using the property of exponents \((a^m)^n = a^{mn}\), we multiply the powers: \( \left(4x^2 - 9y^2\right)^{2 \times 2} = \left(4x^2 - 9y^2\right)^4 \)

So, the simplified expression is \( (4x^2 - 9y^2)^4 \).

Determining the Number of Terms in Expansion

The simplified expression is a binomial raised to the power of 4. A binomial is an algebraic expression with two terms, like \( (4x^2 - 9y^2) \). When a binomial of the form \( (A+B)^n \) or \( (A-B)^n \) is expanded, the number of terms in the expansion is given by \(n+1\).

In our simplified expression \( (4x^2 - 9y^2)^4 \):

The binomial is \( (4x^2 - 9y^2) \).

The power \(n\) is 4.

The number of terms in the expansion is \(n+1 = 4+1 = 5\).

Let's verify this with a simple example:

Expansion of \( (a+b)^2 = a^2 + 2ab + b^2 \). Number of terms = 3. Here \(n=2\), \(n+1=3\).

Expansion of \( (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \). Number of terms = 4. Here \(n=3\), \(n+1=4\).

This pattern holds true for any positive integer power \(n\). Thus, for \( (4x^2 - 9y^2)^4 \), the number of terms will be \(4+1=5\).

Summary of the Calculation

We started with \( \left\{(2x – 3y)^2 (2x + 3y)^2\right\}^2 \) and simplified it as follows:

The final simplified form \( (4x^2 - 9y^2)^4 \) is a binomial \( (A+B)^n \) where \(A = 4x^2\), \(B = -9y^2\), and \(n = 4\). The number of terms in such an expansion is \(n+1 = 4+1=5\).

Revision Table: Key Concepts for Binomial Expansion

Additional Information: More on Binomial Theorem

The Binomial Theorem provides a formula for expanding any binomial \( (a+b)^n \) for any non-negative integer \(n\). The general form is:

\( (a+b)^n = \binom{n}{0} a^n b^0 + \binom{n}{1} a^{n-1} b^1 + \binom{n}{2} a^{n-2} b^2 + \dots + \binom{n}{n-1} a^1 b^{n-1} + \binom{n}{n} a^0 b^n \)

Here, \( \binom{n}{k} \) represents the binomial coefficient, calculated as \( \frac{n!}{k!(n-k)!} \). These coefficients correspond to the numbers in Pascal's Triangle.

The terms in the expansion are of the form \( \binom{n}{k} a^{n-k} b^k \), where \(k\) ranges from 0 to \(n\). Since \(k\) takes on \(n+1\) different values (0, 1, 2, ..., n), there are \(n+1\) terms in the expansion. In our problem, the variable parts of the terms would be combinations of powers of \(x^2\) and \(y^2\), like \((x^2)^4 (y^2)^0\), \((x^2)^3 (y^2)^1\), \((x^2)^2 (y^2)^2\), \((x^2)^1 (y^2)^3\), \((x^2)^0 (y^2)^4\). These simplify to \(x^8\), \(x^6 y^2\), \(x^4 y^4\), \(x^2 y^6\), \(y^8\). Since all these terms have different combinations of powers of \(x\) and \(y\), they are unlike terms and cannot be combined, resulting in exactly 5 terms.