Find n if :
` ""^(2n+1) P_(n-1) : ""^(2n-1) P_n = 3: 5`

Answer

To solve the problem, we need to find the value of \( n \) given the equation:

\[ \frac{(2n+1) P (n-1)}{(2n-1) P n} = \frac{3}{5} \]

### Step 1: Write the Permutation Formula
Recall the permutation formula:
\[ nPr = \frac{n!}{(n-r)!} \]

### Step 2: Apply the Permutation Formula
Using the permutation formula, we can express the left side of the equation:
\[ (2n+1) P (n-1) = \frac{(2n+1)!}{(2n+1 - (n-1))!} = \frac{(2n+1)!}{(n+2)!} \]
\[ (2n-1) P n = \frac{(2n-1)!}{(2n-1 - n)!} = \frac{(2n-1)!}{(n-1)!} \]

### Step 3: Substitute into the Equation
Substituting these back into the equation gives us:
\[ \frac{\frac{(2n+1)!}{(n+2)!}}{\frac{(2n-1)!}{(n-1)!}} = \frac{3}{5} \]

### Step 4: Simplify the Left Side
This simplifies to:
\[ \frac{(2n+1)! \cdot (n-1)!}{(2n-1)! \cdot (n+2)!} = \frac{3}{5} \]

### Step 5: Factorial Simplification
We can express \( (2n+1)! \) in terms of \( (2n-1)! \):
\[ (2n+1)! = (2n+1)(2n)(2n-1)! \]
Thus, we have:
\[ \frac{(2n+1)(2n)(n-1)!}{(2n-1)!(n+2)!} = \frac{3}{5} \]

### Step 6: Further Simplification
This can be rewritten as:
\[ \frac{(2n+1)(2n)}{(n+2)(n+1)} = \frac{3}{5} \]

### Step 7: Cross-Multiply
Cross-multiplying gives:
\[ 5(2n+1)(2n) = 3(n+2)(n+1) \]

### Step 8: Expand Both Sides
Expanding both sides:
\[ 5(4n^2 + 2n) = 3(n^2 + 3n + 2) \]
\[ 20n^2 + 10n = 3n^2 + 9n + 6 \]

### Step 9: Rearranging the Equation
Rearranging gives:
\[ 20n^2 - 3n^2 + 10n - 9n - 6 = 0 \]
\[ 17n^2 + n - 6 = 0 \]

### Step 10: Factor the Quadratic
Now we need to factor or use the quadratic formula to solve for \( n \):
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 17, b = 1, c = -6 \).

### Step 11: Calculate the Discriminant
Calculating the discriminant:
\[ b^2 - 4ac = 1^2 - 4(17)(-6) = 1 + 408 = 409 \]

### Step 12: Solve for \( n \)
Now substituting back into the quadratic formula:
\[ n = \frac{-1 \pm \sqrt{409}}{34} \]

### Step 13: Finding the Value of \( n \)
We can approximate \( \sqrt{409} \) to find the values of \( n \). The possible values will be:
1. \( n_1 = \frac{-1 + \sqrt{409}}{34} \)
2. \( n_2 = \frac{-1 - \sqrt{409}}{34} \) (not valid since \( n \) must be positive)

### Conclusion
The valid solution for \( n \) is:
\[ n \approx \frac{-1 + 20.223}{34} \approx 0.57 \text{ (not an integer)} \]
Thus, we will need to check for integer solutions or further evaluate the quadratic.