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:

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

In our case, we can identify \( a = 1 \), \( b = -x^2 \), and \( n = 10 \). Thus, the expansion of \( (1 - x^2)^{10} \) can be expressed as:

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

This simplifies to:

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

We are interested in finding the coefficient of \( x^{10} \). To find this, we need to determine the value of \( r \) such that \( 2r = 10 \). Solving for \( r \):

\[ 2r = 10 \implies r = 5 \]

Now, we need to find the term corresponding to \( r = 5 \) in the expansion. The term is given by:

\[ \binom{10}{5} (-1)^5 x^{10} \]

The coefficient of \( x^{10} \) in this term is:

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

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

\[ \binom{10}{5} = \frac{10!}{5! \cdot 5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \]

Now, since \( (-1)^5 = -1 \), the coefficient becomes:

\[ 252 \times (-1) = -252 \]

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

\[ \boxed{-252} \]