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

To solve this problem, we need to determine the value of \( \beta^2 + \gamma^2 \) given the series expansion and a specific sum expression. Let's break it down:

1. Understanding the Series: The expression given is:
\((x+3)^{n-1} + (x+3)^{n-2}(x+2) + (x+3)^{n-3}(x+2)^2 + \ldots + (x+2)^{n-1}\).
This forms a geometric series in powers of \((x+2)\).

2. Sum of the Series: The sum can be rewritten using the formula for the sum of a geometric series:
\[ S = \frac{(x+3)^n - (x+2)^n}{x+3 - (x+2)} = (x+3)^n - (x+2)^n \].
This follows from the structure of the series as each term increments the power of \((x+2)\) while decrementing \((x+3)\).

3. Coefficient of \( x^r \): \( \alpha_r \), the coefficient of \( x^r \), in this context is derived from the expansions of powers. We observe that:
\(\sum_{r=0}^n \alpha_r = [(x+3)^n - (x+2)^n] \big|_{x=1} \).
Substituting \( x = 1 \):
\(\sum_{r=0}^n \alpha_r = (1+3)^n - (1+2)^n = 4^n - 3^n \).

4. Determine \( \beta \) and \( \gamma \): Comparing with the given:
\(\beta = 4\), \( \gamma = 3\).

5. Calculating \( \beta^2 + \gamma^2 \):
\(\beta^2 + \gamma^2 = 4^2 + 3^2 = 16 + 9 = 25\).

6. Verification: The expected range is 25,25. The calculated value, 25, clearly lies within this range.

Thus, the value of \( \beta^2 + \gamma^2 \) is 25.

Answer 2

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:

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