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

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:

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