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Polynomial expansion is a crucial concept in algebra. It involves expressing a polynomial raised to a power as a sum of individual terms. For instance, the expansion of \((1 - x^2)^n\) involves breaking down \((1 - x^2)^n\) into a series of terms that include powers of \x\. This is done using the binomial theorem. For the given problem, each term in the series \((1 - x^2) + (1 - x^2)^2 + ... + (1 - x^2)^{10}\) is expanded individually. Each term represents a polynomial, and polynomial expansion helps in identifying specific coefficients, such as the coefficient of \x^4\. By understanding polynomial expansion, we can handle each term more systematically and extract useful information such as specific coefficients.

The binomial theorem offers a practical method to expand binomial expressions. It states that \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \. \] For the problem at hand, we use the binomial theorem to expand \((1 - x^2)^n\). Here, \a = 1\ and \b = -x^2\, which transforms the expansion into: \[ (1 - x^2)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k (x^2)^k \. \] This expansion results in a series of terms where each term is of the form \binom{n}{k} (-1)^k x^{2k}\. This allows us to specifically solve for terms like \x^4\ by setting \2k=4\, leading to \k=2\. The binomial theorem simplifies the process of polynomial expansion, making it easier to find specific coefficients.

Finding the sum of coefficients in a polynomial is another important exercise. In our problem, once we identify the coefficient of \x^4\ in each term of the expanded polynomials, we sum these coefficients. Specifically, for \x^4\, the coefficient in \((1 - x^2)^n\) is \binom{n}{2}(-1)^2\, which simplifies to \binom{n}{2}\. We then sum these coefficients for \ ranging from 2 to 10: \[ \sum_{n=2}^{10} \binom{n}{2}. \] Simplifying this further: \[ \sum_{n=2}^{10} \binom{n}{2} = \frac{1}{2} \sum_{n=2}^{10} n(n-1) = (2 \cdot 1) + (3 \cdot 2) + ... + (10 \cdot 9). \] Adding these gives a final sum of 330, and multiplying by \frac{1}{2}\, we obtain the result 165. This process illustrates how to handle large polynomials by breaking down their sums step-by-step.