Coefficient of \(x^{10}\) in Binomial Expansion

This solution explains how to find the specific term \(x^{10}\)'s coefficient within a complex algebraic expression involving binomial powers. We will break down the expression, simplify it, and then use the binomial theorem to find the required coefficient.

Expression Analysis and Simplification

The given expression is:

\(\left(1-x^2\right)^{20} \left(2-x^2-\frac{1}{x^2}\right)^{-5}\)

Our goal is to find the coefficient of the \(x^{10}\) term after expanding this expression.

Simplifying the Term \(\left(2-x^2-\frac{1}{x^2}\right)^{-5}\)

Let's focus on simplifying the second part of the expression first:

\(\left(2-x^2-\frac{1}{x^2}\right)^{-5}\)

Combine the terms inside the parenthesis using a common denominator \(x^2\):

\(= \left(\frac{2x^2 - x^4 - 1}{x^2}\right)^{-5}\)

Factor out \(-1\) from the numerator:

\(= \left(\frac{-(x^4 - 2x^2 + 1)}{x^2}\right)^{-5}\)

Recognize the perfect square in the numerator \(\left(x^4 - 2x^2 + 1 = (x^2-1)^2\right)\):

\(= \left(\frac{-(x^2-1)^2}{x^2}\right)^{-5}\)

Apply the exponent rule \((a/b)^n = a^n / b^n\) and \((-1)^n\):

\(= (-1)^{-5} \frac{((x^2-1)^2)^{-5}}{(x^2)^{-5}}\)

Since \((-1)^{-5} = -1\) and \((a^m)^n = a^{mn}\):

\(= -1 \cdot \frac{(x^2-1)^{-10}}{x^{-10}}\)

\(= -1 \cdot x^{10} \cdot (x^2-1)^{-10}\)

We can also write \((x^2-1)\) as \(-(1-x^2)\). Therefore:

\((x^2-1)^{-10} = (-(1-x^2))^{-10} = (-1)^{-10} (1-x^2)^{-10} = 1 \cdot (1-x^2)^{-10} = (1-x^2)^{-10}\)

Substituting this back, the simplified second term is:

\(-1 \cdot x^{10} \cdot (1-x^2)^{-10} = -x^{10} (1-x^2)^{-10}\)

Combining Factors for Final Expansion

Now, substitute this simplified term back into the original expression:

\(\text{Expression} = (1-x^2)^{20} \cdot \left(-x^{10} (1-x^2)^{-10}\right)\)

Combine the terms with the base \((1-x^2)\) using the rule \(a^m \cdot a^n = a^{m+n}\):

\(= -x^{10} \cdot (1-x^2)^{20} \cdot (1-x^2)^{-10}\)

\(= -x^{10} \cdot (1-x^2)^{20 - 10}\)

\(= -x^{10} (1-x^2)^{10}\)

So, we need to find the coefficient of \(x^{10}\) in the expansion of \(-x^{10} (1-x^2)^{10}\).

Binomial Expansion and Term Identification

We will use the binomial theorem to expand \((1-x^2)^{10}\). The binomial theorem states:

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

Applying Binomial Theorem to \((1-x^2)^{10}\)

In our case, \(a=1\), \(b=-x^2\), and \(n=10\). Applying the theorem:

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

Simplify the terms:

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

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

Combining with \(-x^{10}\)

Now, multiply the expansion by \(-x^{10}\):

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

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

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

Identifying the \(x^{10}\) Term

We are looking for the term where the exponent of \(x\) is 10. So, we set the exponent \(10+2k\) equal to 10:

\(10 + 2k = 10\)

Solve for \(k\):

\(2k = 10 - 10\)

\(2k = 0\)

\(k = 0\)

This value of \(k=0\) falls within the summation range (\(0\) to \(10\)).

Final Coefficient Calculation

To find the coefficient of \(x^{10}\), substitute \(k=0\) into the coefficient part of the summation term:

\(\text{Coefficient} = - \binom{10}{k} (-1)^k \quad \text{when } k=0\)

\(= - \binom{10}{0} (-1)^0\)

Calculate the binomial coefficient and the power of -1:

• \(\binom{10}{0} = 1\)
• \((-1)^0 = 1\)

Substitute these values back:

\(\text{Coefficient} = - (1) \cdot (1)\)

\(= -1\)

Therefore, the coefficient of \(x^{10}\) in the expansion of the given expression is -1.