`P_n=alpha^n+beta^n , alpha +beta=1 , alpha*beta=-1 , P_(n-1)=11 ,P_(n+1)=29 , then P_(n)^2=`

Answer

To solve the problem, we are given the following information:

1. \( P_n = \alpha^n + \beta^n \)
2. \( \alpha + \beta = 1 \)
3. \( \alpha \beta = -1 \)
4. \( P_{n-1} = 11 \)
5. \( P_{n+1} = 29 \)

We need to find \( P_n^2 \).

### Step-by-Step Solution:

**Step 1: Establish the recurrence relation for \( P_n \)**

Using the properties of roots, we can derive a recurrence relation for \( P_n \):
\[ P_n = (\alpha + \beta) P_{n-1} - \alpha \beta P_{n-2} \]
Substituting the known values:
\[ P_n = 1 \cdot P_{n-1} - (-1) \cdot P_{n-2} \]
This simplifies to:
\[ P_n = P_{n-1} + P_{n-2} \]

**Step 2: Use the given values to find \( P_{n-2} \)**

We know:
\[ P_{n-1} = 11 \quad \text{and} \quad P_{n+1} = 29 \]
Using the recurrence relation:
\[ P_{n+1} = P_n + P_n - P_{n-1} \]
So,
\[ 29 = P_n + 11 \]
This gives us:
\[ P_n = 29 - 11 = 18 \]

**Step 3: Calculate \( P_n^2 \)**

Now that we have \( P_n = 18 \), we can find \( P_n^2 \):
\[ P_n^2 = 18^2 = 324 \]

### Final Answer:
\[ P_n^2 = 324 \]