The `(p+2)^(th)` term from the end in `(x-1/x)^(2n+1)` is :

Options

A. `(-1)^(p)((2n+1)!)/((2n-p)!(p+1)!)x^(2p-2n+1)`
B. `(-1)^(p)((2n+1)!)/((2n-p)!(p+1)!)x^(2n-2p+1)`
C. `(-1)^(p)((2n+1)!)/((2n-p)!(p+1)!)x^(2p-2n-1)`
D. None of these

Answer

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

### Step 1: Identify the general term
The general term \(T_r\) in the expansion of \((x - \frac{1}{x})^{2n + 1}\) can be given by the binomial theorem:
\[ T_r = \binom{n}{r} (x)^{2n + 1 - r} \left(-\frac{1}{x}\right)^{r} \]
This simplifies to:
\[ T_r = \binom{2n + 1}{r} x^{(2n + 1 - r) - r} (-1)^r = \binom{2n + 1}{r} (-1)^r x^{2n + 1 - 2r} \]

### Step 2: Determine the term from the end
The \((p + 2)^{th}\) term from the end corresponds to the \((2n + 2 - (p + 2))^{th}\) term from the beginning. Therefore, we need to find:
\[ r = 2n + 2 - (p + 2) = 2n - p \]

### Step 3: Substitute \(r\) into the general term
Now, substituting \(r = 2n - p\) into the general term:
\[ T_{2n - p} = \binom{2n + 1}{2n - p} (-1)^{2n - p} x^{2n + 1 - 2(2n - p)} \]
This simplifies to:
\[ T_{2n - p} = \binom{2n + 1}{2n - p} (-1)^{2n - p} x^{2p + 1 - 2n} \]

### Step 4: Simplify the expression
Since \((-1)^{2n - p} = (-1)^{2n} \cdot (-1)^{-p} = (-1)^{-p} = (-1)^{p}\) (as \((-1)^{2n} = 1\)):
\[ T_{2n - p} = \binom{2n + 1}{p} (-1)^{p} x^{2p + 1 - 2n} \]

### Final Result
Thus, the \((p + 2)^{th}\) term from the end in \((x - \frac{1}{x})^{2n + 1}\) is:
\[ T_{p + 2} = \binom{2n + 1}{p} (-1)^{p} x^{2p + 1 - 2n} \]