Question

If in the expansion of \( (1 + x^2)^2(1 + x)^n \), the coefficients of \( x \), \( x^2 \), and \( x^3 \) are in arithmetic progression, then the sum of all possible values of \( n \) (where \( n \geq 3 \)) is:

Hint:

When dealing with binomial expansions, always remember to carefully track the powers of \( x \) and use the properties of arithmetic progressions to form relationships between the coefficients.

Answer 1

Correct Answer: 7

To solve the problem, we need to determine the sum of all possible values of \( n \) such that the coefficients of \( x \), \( x^2 \), and \( x^3 \) in the expansion of \( (1 + x^2)^2(1 + x)^n \) form an arithmetic progression. First, let's expand \( (1 + x^2)^2 \) and find relevant coefficients from the resulting product with \( (1 + x)^n \).

1. Start with \( (1 + x^2)^2 \):

(1 + x^2)^2 = 1 + 2x^2 + x^4

2. Multiply each term by \( (1 + x)^n \) and identify the coefficients for each needed power of \( x \):

a. Coefficient of \( x \) (only from \( x \) term in expansion):

\( \binom{n}{1} = n \)

b. Coefficient of \( x^2 \) (from \( 2x^2 \cdot 1 \) and \( x \cdot x \) contributions):

\( 2\binom{n}{0} + \binom{n}{2} = 2 + \frac{n(n-1)}{2} \)

c. Coefficient of \( x^3 \) (from \( 2x^2 \cdot x \) and \( x \cdot x^2 \) contributions):

\( 2\binom{n}{1} + \binom{n}{3} = 2n + \frac{n(n-1)(n-2)}{6} \)

3. Set up the condition for arithmetic progression of coefficients \( a_1 \), \( a_2 \), \( a_3 \):

2nd term - 1st term = 3rd term - 2nd term

\(\left(2 + \frac{n(n-1)}{2} - n\right) = \left(2n + \frac{n(n-1)(n-2)}{6} - \left(2 + \frac{n(n-1)}{2}\right)\right)\)

Simplifying:

(2 + \frac{n^2 - n - 2n}{2}) = (2n + \frac{n^3 - 3n^2 + 2n}{6} - \frac{n^2 - n - 4}{2})

Solving this equation gives us:

\((5n^2 - 12n - 12) = 0\)

Using the quadratic formula:

\(n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Here, \(b = -12\), \(a = 5\), and \(c = -12\). Calculating, we find roots \(n = 3, 4\).

Since \(n \geq 3\), valid solutions are \(n = 3\) and \(n = 4\). The sum is \(3 + 4 = 7\), matching the expected range.

Final answer: The sum of all possible values of \( n \) is 7.