Question

The sum of all possible values of \( n \in \mathbb{N} \), so that the coefficients of \(x, x^2\) and \(x^3\) in the expansion of \((1+x^2)^2(1+x)^n\) are in arithmetic progression is :

• A. 9
• B. 3
• C. 7
• D. 12

Hint:

For small \(n\), you can quickly test values. For \(n=3\), coeffs are \(3, 5, 7\), which are clearly in AP with a common difference of 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Expand \((1 + 2x^2 + x^4)(1 + x)^n\). Collect the terms for \(x, x^2,\) and \(x^3\). Use the AP condition: \(2 \times (\text{Coeff. of } x^2) = (\text{Coeff. of } x) + (\text{Coeff. of } x^3)\).
Step 2: Key Formula or Approach:
General term of \((1+x)^n\) is \(^nC_r x^r\).
Step 3: Detailed Explanation:
- Coeff. of \(x\): \(^nC_1 = n\). - Coeff. of \(x^2\): \(^nC_2 + 2(^nC_0) = \frac{n(n-1)}{2} + 2\). - Coeff. of \(x^3\): \(^nC_3 + 2(^nC_1) = \frac{n(n-1)(n-2)}{6} + 2n\). Applying \(2a_2 = a_1 + a_3\): \[ 2\left(\frac{n^2-n+4}{2}\right) = n + \frac{n^3-3n^2+2n}{6} + 2n \] \[ n^2-n+4 = 3n + \frac{n^3-3n^2+2n}{6} \] Solving the resulting cubic equation for \(n \in \mathbb{N}\) yields \(n=3\).
Step 4: Final Answer:
Sum of values of \(n\) is 3.