1. Understanding the Problem and Key Concepts

The problem asks us to find the value of $\beta^2 + \gamma^2$, given an expression for the sum of coefficients of $x^r$ in a specific polynomial expansion. The polynomial is presented as a sum of terms. The final form of the sum of coefficients is given as $\beta^n - \gamma^n$.

The key concept here involves two fundamental ideas:

• Sum of Coefficients: For any polynomial $P(x) = a_0 + a_1 x + a_2 x^2 + \ldots + a_n x^n$, the sum of all its coefficients ($\sum_{r=0}^n a_r$) can be found by simply substituting $x=1$ into the polynomial, i.e., $\sum_{r=0}^n a_r = P(1)$. This works because when $x=1$, each $x^r$ term becomes $1$, leaving only its coefficient.
• Geometric Progression (GP) Sum: A series where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. The sum of the first $N$ terms of a GP is given by the formula: $S_N = a \frac{1 - r^N}{1 - r}$ where $a$ is the first term and $r$ is the common ratio ($r \neq 1$).

2. Calculating the Sum of Coefficients ($\sum \alpha_r$)

Let the given expression be $P(x)$: $P(x) = (x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots \ldots \ldots .+(x+2)^{n-1}$ The problem states that $\alpha_r$ is the coefficient of $x^r$ in the expansion of $P(x)$. Therefore, the sum of all coefficients, $\sum_{r=0}^n \alpha_r$, is obtained by setting $x=1$ in $P(x)$.

Let's substitute $x=1$ into each term of the series: $\sum_{r=0}^n \alpha_r = (1+3)^{n-1} + (1+3)^{n-2}(1+2) + (1+3)^{n-3}(1+2)^2 + \ldots \ldots \ldots . + (1+2)^{n-1}$ $\sum_{r=0}^n \alpha_r = 4^{n-1} + 4^{n-2} \cdot 3 + 4^{n-3} \cdot 3^2 + \ldots \ldots \ldots . + 3^{n-1}$

3. Recognizing and Summing the Geometric Progression

Now we have a new series: $S = 4^{n-1} + 4^{n-2} \cdot 3 + 4^{n-3} \cdot 3^2 + \ldots \ldots \ldots . + 3^{n-1}$ We observe that this is a Geometric Progression.

• The first term ($a$) is $4^{n-1}$.
• To find the common ratio ($r$), we divide the second term by the first term (or any term by its preceding term): $r = \frac{4^{n-2} \cdot 3}{4^{n-1}} = \frac{3}{4}$
• The number of terms ($N$) in the series can be determined by looking at the powers. The powers of $(x+2)$ (or $3$ after substitution) range from $0$ (in the first term, $(x+2)^0$) to $n-1$ (in the last term, $(x+2)^{n-1}$). This indicates there are $n-1 - 0 + 1 = n$ terms.

Now, we apply the GP sum formula $S_N = a \frac{1 - r^N}{1 - r}$: $\sum_{r=0}^n \alpha_r = 4^{n-1} \cdot \frac{1 - \left(\frac{3}{4}\right)^n}{1 - \frac{3}{4}}$

4. Simplifying the Expression

Let's simplify the sum: $\sum_{r=0}^n \alpha_r = 4^{n-1} \cdot \frac{1 - \frac{3^n}{4^n}}{\frac{4-3}{4}}$ $\sum_{r=0}^n \alpha_r = 4^{n-1} \cdot \frac{\frac{4^n - 3^n}{4^n}}{\frac{1}{4}}$ To simplify further, we can multiply the numerator by the reciprocal of the denominator: $\sum_{r=0}^n \alpha_r = 4^{n-1} \cdot \frac{4^n - 3^n}{4^n} \cdot 4$ $\sum_{r=0}^n \alpha_r = \frac{4^{n-1} \cdot 4 \cdot (4^n - 3^n)}{4^n}$ Using the exponent rule $a^p \cdot a^q = a^{p+q}$, we have $4^{n-1} \cdot 4 = 4^{(n-1)+1} = 4^n$: $\sum_{r=0}^n \alpha_r = \frac{4^n \cdot (4^n - 3^n)}{4^n}$ We can cancel out $4^n$ from the numerator and denominator: $\sum_{r=0}^n \alpha_r = 4^n - 3^n$

5. Determining $\beta$ and $\gamma$

The problem states that $\sum_{r=0}^n \alpha_r = \beta^n - \gamma^n$. By comparing our derived sum with the given form: $4^n - 3^n = \beta^n - \gamma^n$ Given that $\beta, \gamma \in \mathbb{N}$ (natural numbers), we can directly infer: $\beta = 4$ $\gamma = 3$

6. Calculating the Final Value

Finally, we need to find the value of $\beta^2 + \gamma^2$: $\beta^2 + \gamma^2 = 4^2 + 3^2$ $\beta^2 + \gamma^2 = 16 + 9$ $\beta^2 + \gamma^2 = 25$

7. Tips and Common Mistakes

• Tip 1: Identify the pattern early. When faced with a sum of terms, always check if it follows an Arithmetic Progression (AP), Geometric Progression (GP), or another common series. This significantly simplifies the problem.
• Tip 2: Sum of coefficients is $P(1)$. This is a powerful shortcut. Do not try to expand the entire polynomial first to find coefficients unless explicitly asked.
• Common Mistake 1: Incorrectly counting the number of terms in a GP. Pay close attention to the starting and ending powers to ensure you count $N$ correctly. In this case, powers from $0$ to $n-1$ means $n$ terms.
• Common Mistake 2: Algebraic errors during simplification. Be careful with exponent rules and fraction arithmetic.

8. Summary and Key Takeaway

This problem effectively tests the understanding and application of two crucial concepts: finding the sum of coefficients of a polynomial and summing a geometric progression. By first substituting $x=1$ into the given series, we transformed the problem into summing a simpler GP. This strategic simplification led directly to the values of $\beta$ and $\gamma$, allowing us to calculate the final required expression. The key takeaway is to simplify expressions as much as possible at each step and recognize underlying mathematical structures like GPs.