Problem 51 - Algebra and Trigonometry

Question

Find the coefficient of \(x^{4}\) in \(f(x)=\left(1-x^{2}\right)+\left(1-x^{2}\right)^{2}+\cdots+\left(1-x^{2}\right)^{10}\).

Step-by-Step Solution

Answer

165

Step 1 - Identify the function components

The given function is a sum of terms: \[ f(x) = \left(1 - x^2\right) + \left(1 - x^2\right)^2 + \left(1 - x^2\right)^3 + \cdots + \left(1 - x^2\right)^{10} \].

Step 2 - Expand binomial terms

Each term is a binomial expansion: \[ \left(1 - x^2\right)^n = \sum_{k=0}^{n} \binom{n}{k} (-1)^k (x^2)^k = \sum_{k=0}^{n} \binom{n}{k} (-1)^k x^{2k}. \] We need to find the coefficients of \(x^4\) in each expansion.

Step 3 - Find specific coefficients

We are looking for the coefficients of \(x^4\), meaning in each term \(\left(1 - x^2\right)^n\), we identify the coefficient where \(2k = 4\). Thus, \(k = 2\). The coefficient for \(x^4\) in \(\left(1 - x^2\right)^n\) is given by: \[ \binom{n}{2}(-1)^2 = \binom{n}{2}. \]

Step 4 - Sum coefficients from all terms

Now sum these coefficients from \(n = 2\) to \(n = 10\), since for \(n=0\) and \(n=1\), there are no \(x^4\) terms:\[ \sum_{n=2}^{10} \binom{n}{2}. \]

Step 5 - Simplify final sum

The sum of binomial coefficients can be simplified as:\[ \sum_{n=2}^{10} \binom{n}{2} = \sum_{n=2}^{10} \frac{n(n-1)}{2} = \frac{1}{2} \sum_{n=2}^{10} n(n-1). \] This simplifies further to:\[ \sum_{n=2}^{10} n(n-1) = (2 \cdot 1) + (3 \cdot 2) + (4 \cdot 3) + (5 \cdot 4) + (6 \cdot 5) + (7 \cdot 6) + (8 \cdot 7) + (9 \cdot 8) + (10 \cdot 9). \]

Step 6 - Calculate the total sum

Calculate the sum:\[ 2 + 6 + 12 + 20 + 30 + 42 + 56 + 72 + 90 = 330. \] Then multiply by \(\frac{1}{2}\) to get the final coefficient sum: \[ \frac{1}{2} \times 330 = 165. \]

Key Concepts

Polynomial Expansion

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.

Binomial Theorem

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.

Coefficient Sum

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.