Find the coefficient of `x^(10)` in the expansion of `(1-x^(2))^(10)`

Explanation

To find the coefficient of \( x^{10} \) in the expansion of \( (1 - x^2)^{10} \), we can use the Binomial Theorem. The Binomial Theorem states that:

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

In our case, we can rewrite the expression as:

\[ (1 - x^2)^{10} = \sum_{k=0}^{10} \binom{10}{k} (1)^{10-k} (-x^2)^k \]

This simplifies to:

\[ (1 - x^2)^{10} = \sum_{k=0}^{10} \binom{10}{k} (-1)^k x^{2k} \]

Now, we need to find the coefficient of \( x^{10} \) in this expansion. To do this, we set the exponent of \( x \) equal to 10:

\[ 2k = 10 \]

Solving for \( k \):

\[ k = \frac{10}{2} = 5 \]

Now, we need to find the coefficient corresponding to \( k = 5 \):

\[ \text{Coefficient} = \binom{10}{5} (-1)^5 \]

Calculating \( \binom{10}{5} \):

\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} \]

Calculating this step-by-step:

1. Calculate the numerator:
\[ 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
\[ 5040 \times 6 = 30240 \]

2. Calculate the denominator:
\[ 5 \times 4 = 20 \]
\[ 20 \times 3 = 60 \]
\[ 60 \times 2 = 120 \]
\[ 120 \times 1 = 120 \]

3. Now divide the numerator by the denominator:
\[ \frac{30240}{120} = 252 \]

Thus, we have:

\[ \text{Coefficient} = \binom{10}{5} (-1)^5 = 252 \cdot (-1) = -252 \]

Therefore, the coefficient of \( x^{10} \) in the expansion of \( (1 - x^2)^{10} \) is:

\[ \boxed{-252} \]