Let `(1+x^2)^2 . (1+x)^n = Sigma_(k=0)^(n+4) (a_k).x ^k "if" a_1,a_2 & a_3` are in AP find n

Answer

To solve the problem, we need to analyze the expression given and find the value of \( n \) such that \( a_1, a_2, a_3 \) are in Arithmetic Progression (AP).

### Step-by-Step Solution:

1. **Expand the Expression:**
We start with the expression \((1 + x^2)^2 \cdot (1 + x)^n\).
\[ (1 + x^2)^2 = 1 + 2x^2 + x^4 \]
Therefore, the expression becomes:
\[ (1 + 2x^2 + x^4)(1 + x)^n \]

2. **Distribute the Terms:**
We will distribute \((1 + 2x^2 + x^4)\) with \((1 + x)^n\):
\[ (1)(1 + x)^n + (2x^2)(1 + x)^n + (x^4)(1 + x)^n \]
This gives us:
\[ (1 + x)^n + 2x^2(1 + x)^n + x^4(1 + x)^n \]

3. **Identify Coefficients:**
The coefficients \( a_k \) in the expansion can be identified as:
- \( a_1 = \text{coefficient of } x^1 \)
- \( a_2 = \text{coefficient of } x^2 \)
- \( a_3 = \text{coefficient of } x^3 \)

From the expansion:
- \( a_1 = \binom{n}{1} = n \)
- \( a_2 = \binom{n}{2} + 2 \) (from \( 2x^2 \))
- \( a_3 = \binom{n}{3} + 2\binom{n}{1} = \binom{n}{3} + 2n \)

4. **Set Up the AP Condition:**
For \( a_1, a_2, a_3 \) to be in AP:
\[ 2a_2 = a_1 + a_3 \]
Substituting the values:
\[ 2\left(\binom{n}{2} + 2\right) = n + \left(\binom{n}{3} + 2n\right) \]
Simplifying gives:
\[ 2\binom{n}{2} + 4 = 3n + \binom{n}{3} \]

5. **Substitute Binomial Coefficients:**
Recall that:
\[ \binom{n}{2} = \frac{n(n-1)}{2}, \quad \binom{n}{3} = \frac{n(n-1)(n-2)}{6} \]
Substitute these into the equation:
\[ 2\left(\frac{n(n-1)}{2}\right) + 4 = 3n + \frac{n(n-1)(n-2)}{6} \]
This simplifies to:
\[ n(n-1) + 4 = 3n + \frac{n(n-1)(n-2)}{6} \]

6. **Clear the Fraction:**
Multiply through by 6 to eliminate the fraction:
\[ 6n(n-1) + 24 = 18n + n(n-1)(n-2) \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ n^3 - 9n^2 + 26n - 24 = 0 \]

8. **Finding Roots:**
We can use the Rational Root Theorem or synthetic division to find the roots of the cubic equation. Testing \( n = 9 \):
\[ 9^3 - 9(9^2) + 26(9) - 24 = 729 - 729 + 234 - 24 = 0 \]
Thus, \( n = 9 \) is a root.

### Final Answer:
The value of \( n \) is \( 9 \).