Question

Let the coefficient of \( x^r \) in the expansion of \((x + 3)^{n-1} + (x + 3)^{n-2} (x + 2) + (x + 3)^{n-3} (x + 2)^2 + \ldots + (x + 2)^{n-1}\)
be \( \alpha_r \). If \( \sum_{r=0}^n \alpha_r = \beta^n - \gamma^n \), \( \beta, \gamma \in \mathbb{N} \), then the value of \( \beta^2 + \gamma^2 \) equals \(\_\_\_\_\_\).

Answer 1

Correct Answer: 25

Consider the expansion:

\[ (x + 3)^{n-1} + (x + 3)^{n-2}(x + 2) + (x + 3)^{n-3}(x + 2)^2 + \ldots + (x + 2)^{n-1} \]

The sum of coefficients \(\sum_{r=0}^{n} \alpha_r\) is given by:

\[ \sum_{r=0}^{n} \alpha_r = 4^{n-1} + 4^{n-2} \times 3 + 4^{n-3} \times 3^2 + \ldots + 3^{n-1} \]

This forms a geometric series with the first term \(4^{n-1}\) and common ratio \(\frac{3}{4}\):

\[ \sum_{r=0}^{n} \alpha_r = 4^{n-1} \left(1 + \frac{3}{4} + \left(\frac{3}{4}\right)^2 + \ldots + \left(\frac{3}{4}\right)^{n-1}\right) \]

The sum of the geometric series is given by:

\[ \sum_{r=0}^{n} \alpha_r = 4^{n-1} \times \frac{1 - \left(\frac{3}{4}\right)^n}{1 - \frac{3}{4}} = 4^n - 3^n \]

Given:

\[ \sum_{r=0}^{n} \alpha_r = \beta^n - \gamma^n \]

Comparing:

\[ \beta = 4, \quad \gamma = 3 \]

The value of \(\beta^2 + \gamma^2\) is:

\[ \beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25 \]