Understanding the Binomial Expansion Problem

We are asked to find the number of terms in the expansion of the expression \(\left(1 + \frac{2}{x}\right)^9\left(1-\frac{2}{x}\right)^9\). This expression involves powers of binomials.

Simplifying the Expression

Let's first simplify the given expression. The expression is in the form \((a+b)^n (a-b)^n\), where \(a = 1\), \(b = \frac{2}{x}\), and \(n = 9\).

We know the algebraic identity \((a+b)(a-b) = a^2 - b^2\). Using this identity, we can simplify the product of the two terms inside the parentheses first:

$$ \left(1 + \frac{2}{x}\right)\left(1 - \frac{2}{x}\right) = 1^2 - \left(\frac{2}{x}\right)^2 = 1 - \frac{4}{x^2} $$

Now, the original expression can be rewritten as:

$$ \left(1 + \frac{2}{x}\right)^9\left(1 - \frac{2}{x}\right)^9 = \left[\left(1 + \frac{2}{x}\right)\left(1 - \frac{2}{x}\right)\right]^9 = \left(1 - \frac{4}{x^2}\right)^9 $$

So, the problem reduces to finding the number of terms in the expansion of \(\left(1 - \frac{4}{x^2}\right)^9\).

Applying the Binomial Theorem

The expression \(\left(1 - \frac{4}{x^2}\right)^9\) is a binomial in the form \((A+B)^n\), where \(A=1\), \(B=-\frac{4}{x^2}\), and \(n=9\).

According to the Binomial Theorem, the expansion of \((A+B)^n\) is given by:

$$ (A+B)^n = \binom{n}{0}A^n B^0 + \binom{n}{1}A^{n-1} B^1 + \binom{n}{2}A^{n-2} B^2 + \dots + \binom{n}{n}A^0 B^n $$

Each term in this sum corresponds to a different value of \(k\) from 0 to \(n\) in the general term \(\binom{n}{k}A^{n-k} B^k\).

The terms in the expansion of \((A+B)^n\) are generated for \(k = 0, 1, 2, \dots, n\). The number of possible values for \(k\) is \(n - 0 + 1 = n+1\).

Therefore, a binomial expansion of the form \((A+B)^n\) has exactly \(n+1\) terms, provided that \(A\) and \(B\) are not single terms that simplify the number of terms further (e.g., if \(A\) and \(B\) are constants or simple algebraic terms like in our case, where terms with different powers of \(x\) will not combine).

Calculating the Number of Terms

In our simplified expression \(\left(1 - \frac{4}{x^2}\right)^9\), the power is \(n=9\).

Using the property that the expansion of \((A+B)^n\) has \(n+1\) terms, the number of terms in the expansion of \(\left(1 - \frac{4}{x^2}\right)^9\) is \(9+1\).

Number of terms = \(9 + 1 = 10\).

The terms in the expansion will involve powers of \(\left(-\frac{4}{x^2}\right)\), specifically \(\left(-\frac{4}{x^2}\right)^k\) for \(k=0, 1, \dots, 9\). The powers of \(x\) will be \(x^0, x^{-2}, x^{-4}, \dots, x^{-18}\). These are all distinct powers of \(x\), so there will be no combining of terms.

Conclusion on Number of Terms

The simplified expression is \(\left(1 - \frac{4}{x^2}\right)^9\). This is a binomial raised to the power of 9. The number of terms in the expansion of a binomial \((a+b)^n\) is \(n+1\). Since \(n=9\), the number of terms is \(9+1=10\).

Revision Table: Binomial Expansion Terms

Additional Information: General Binomial Expansion

The general form of the binomial expansion for \((a+b)^n\) is:

$$ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k $$

where \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\) are the binomial coefficients. The sum runs from \(k=0\) to \(k=n\), which gives \(n+1\) terms. Each term is of the form \(\binom{n}{k} a^{n-k} b^k\).

For example, the expansion of \((x+y)^3\) is:

• \(k=0\): \(\binom{3}{0}x^{3-0}y^0 = 1 \cdot x^3 \cdot 1 = x^3\)
• \(k=1\): \(\binom{3}{1}x^{3-1}y^1 = 3 \cdot x^2 \cdot y = 3x^2y\)
• \(k=2\): \(\binom{3}{2}x^{3-2}y^2 = 3 \cdot x^1 \cdot y^2 = 3xy^2\)
• \(k=3\): \(\binom{3}{3}x^{3-3}y^3 = 1 \cdot x^0 \cdot y^3 = y^3\)

The expansion is \(x^3 + 3x^2y + 3xy^2 + y^3\), which has \(3+1=4\) terms. This confirms the \(n+1\) rule for the number of terms in a binomial expansion.