(p + 2)th term from the end in the binomial expansion of `(x^(2)-2/(x^(2)))^(2n+1)` is

Options

A. `""^(2n+1)C_(2n-p)(-2)^(2n-p)x^(2p+1-2n)`
B. `""^(2n+1)C_(2n-p)(-2)^(2n-p)x^(2n-2p)`
C. `""^(2n+1)C_(2n-p)(-2)^(2n-p)x^(2n-2p+1)`
D. `""^(2n+1)C_(2n-p)(-2)^(2n-p)x^(2p-1+2n)`

Answer

To find the \((p + 2)\)th term from the end in the binomial expansion of \((x^2 - \frac{2}{x^2})^{2n + 1}\), we will follow these steps:

### Step 1: Understand the Binomial Expansion
The general term in the binomial expansion of \((a + b)^n\) is given by:
\[ T_r = \binom{n}{r} a^{n-r} b^r \]
In our case, \(a = x^2\), \(b = -\frac{2}{x^2}\), and \(n = 2n + 1\).

### Step 2: Identify the Total Number of Terms
In the expansion of \((x^2 - \frac{2}{x^2})^{2n + 1}\), the total number of terms is \(n + 1 = (2n + 1) + 1 = 2n + 2\).

### Step 3: Find the Corresponding Term from the Beginning
The \((p + 2)\)th term from the end corresponds to the \((2n + 2 - (p + 2))\)th term from the beginning, which simplifies to:
\[ (2n + 2 - p - 2) = (2n - p + 1) \]
Thus, we need to find the \((2n - p + 1)\)th term from the beginning.

### Step 4: Substitute into the General Term Formula
Using the general term formula:
\[ T_{r} = \binom{n}{r} a^{n-r} b^r \]
For our case:
- \(n = 2n + 1\)
- \(r = 2n - p\)

Thus, the term we need is:
\[ T_{2n - p + 1} = \binom{2n + 1}{2n - p} (x^2)^{(2n + 1) - (2n - p)} \left(-\frac{2}{x^2}\right)^{2n - p} \]

### Step 5: Simplify the Expression
Calculating the powers:
- The exponent of \(x^2\) becomes:
\[ (2n + 1) - (2n - p) = p + 1 \]
- The exponent of \(-\frac{2}{x^2}\) becomes:
\[ 2n - p \]

Thus, we have:
\[ T_{2n - p + 1} = \binom{2n + 1}{2n - p} (x^2)^{p + 1} \left(-\frac{2}{x^2}\right)^{2n - p} \]

### Step 6: Combine the Terms
Now, we can combine the terms:
\[ T_{2n - p + 1} = \binom{2n + 1}{2n - p} (x^{2(p + 1)}) \left(-2\right)^{2n - p} \left(\frac{1}{x^{2(2n - p)}}\right) \]
This simplifies to:
\[ = \binom{2n + 1}{2n - p} (-2)^{2n - p} x^{2(p + 1) - 2(2n - p)} \]
\[ = \binom{2n + 1}{2n - p} (-2)^{2n - p} x^{4p - 4n + 2} \]

### Final Answer
Thus, the \((p + 2)\)th term from the end in the binomial expansion of \((x^2 - \frac{2}{x^2})^{2n + 1}\) is:
\[ \binom{2n + 1}{2n - p} (-2)^{2n - p} x^{4p - 4n + 2} \]