Finding the Middle Term in the Expansion

We are asked to find the middle term in the expansion of \((x^2 + \frac{1}{x^2} + 2)^n\).

First, let's simplify the expression inside the parenthesis. We can observe that \(x^2 + \frac{1}{x^2} + 2\) is a perfect square. It can be written as \((x + \frac{1}{x})^2\).

So, the given expression becomes:

\((x^2 + \frac{1}{x^2} + 2)^n = \left(\left(x + \frac{1}{x}\right)^2\right)^n = \left(x + \frac{1}{x}\right)^{2n}\)

Now, we need to find the middle term in the expansion of \(\left(x + \frac{1}{x}\right)^{2n}\). This is a binomial expansion of the form \((a+b)^N\), where \(a=x\), \(b=\frac{1}{x}\), and the power \(N=2n\).

The total number of terms in the expansion of \((a+b)^N\) is \(N+1\). In this case, the total number of terms is \(2n + 1\).

Since the total number of terms (\(2n+1\)) is an odd number, there is only one middle term.

The position of the middle term in an expansion with \(N+1\) terms (where \(N\) is even) is given by \(\frac{N}{2} + 1\).

Here, the power is \(N = 2n\), which is an even number. Therefore, the position of the middle term is:

Position of middle term = \(\frac{2n}{2} + 1 = n + 1\)

The general term in the binomial expansion of \((a+b)^N\) is given by \(T_{r+1} = \binom{N}{r} a^{N-r} b^r\).

For the middle term, the position is \(n+1\). This means \(r+1 = n+1\), so \(r = n\). We have \(N=2n\), \(a=x\), and \(b=\frac{1}{x}\).

Substitute these values into the general term formula to find the middle term \(T_{n+1}\):

\(T_{n+1} = \binom{2n}{n} (x)^{2n-n} \left(\frac{1}{x}\right)^n\)

\(T_{n+1} = \binom{2n}{n} x^n \frac{1}{x^n}\)

\(T_{n+1} = \binom{2n}{n} x^n \cdot x^{-n}\)

\(T_{n+1} = \binom{2n}{n} x^{n-n}\)

\(T_{n+1} = \binom{2n}{n} x^0\)

\(T_{n+1} = \binom{2n}{n} \cdot 1\)

\(T_{n+1} = \binom{2n}{n}\)

Now, we express the binomial coefficient \(\binom{2n}{n}\) using factorials. The formula for \(\binom{N}{r}\) is \(\frac{N!}{r!(N-r)!}\).

So, \(\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!} = \frac{(2n)!}{n!n!} = \frac{(2n)!}{(n!)^2}\).

Therefore, the middle term in the expansion of \((x^2 + \frac{1}{x^2} + 2)^n\) is \(\frac{(2n)!}{(n!)^2}\).

Let's compare this result with the given options:

• Option 1: \(\frac{(2n)!}{(n!)^2}\) - This matches our calculated middle term.
• Option 2: \(\frac{(2n!)}{[(\frac{n}{2})!]^2}\) - This does not match.
• Option 3: \(\frac{n}{[(\frac{n}{2})!]^2}\) - This does not match.
• Option 4: \(\frac{1.2.5..(2n + 1)}{n!}2^n\) - This does not match.

The middle term is \(\frac{(2n)!}{(n!)^2}\).