Key Concept: The Binomial Theorem

This problem fundamentally relies on the Binomial Theorem, which provides a formula for expanding powers of binomials. For any non-negative integer $n$ and any real numbers $a$ and $b$, the theorem states: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ A particularly useful form for this problem is when $a=1$: $(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + \binom{n}{n}x^n$ By substituting $\binom{n}{0} = 1$ and $\binom{n}{1} = n$, this expansion becomes: $(1+x)^n = 1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + x^n$ We will use this form to expand $9^n$ as $(1+8)^n$ and $6^n$ as $(1+5)^n$.

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Part 1: Deriving the expression for $\alpha$

We are given the equation: $9^n - 8n - 1 = 64\alpha$

Step 1.1: Express $9^n$ in binomial form To apply the Binomial Theorem, we rewrite $9^n$ as $(1+8)^n$. This form is strategic because $64 = 8^2$, and expanding $(1+8)^n$ will yield terms involving powers of 8, which can then be simplified with the $64$ in the denominator. $9^n = (1+8)^n$

Step 1.2: Expand $(1+8)^n$ using the Binomial Theorem Using the expansion $(1+x)^n = 1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + x^n$ with $x=8$: $(1+8)^n = \binom{n}{0} + \binom{n}{1}(8) + \binom{n}{2}(8)^2 + \binom{n}{3}(8)^3 + \dots + \binom{n}{n}(8)^n$ Substituting the values of the first two binomial coefficients ($\binom{n}{0}=1$ and $\binom{n}{1}=n$): $(1+8)^n = 1 + 8n + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \dots + \binom{n}{n}8^n$

Step 1.3: Substitute the expansion and simplify for $\alpha$ Now, substitute this expanded form of $(1+8)^n$ back into the original equation for $\alpha$: $64\alpha = \left( 1 + 8n + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \dots + \binom{n}{n}8^n \right) - 8n - 1$ We observe that the terms $1$ and $8n$ from the expansion cancel out with the $-8n-1$ terms. This cancellation is crucial and simplifies the expression significantly. $64\alpha = (1-1) + (8n-8n) + \binom{n}{2}8^2 + \binom{n}{3}8^3 + \dots + \binom{n}{n}8^n$ $64\alpha = \binom{n}{2}8^2 + \binom{n}{3}8^3 + \binom{n}{4}8^4 + \dots + \binom{n}{n}8^n$ Finally, divide both sides by $64$ (which is $8^2$) to find $\alpha$. Each term in the summation is divided by $8^2$, reducing the power of 8 by two for each term: $\alpha = \frac{\binom{n}{2}8^2 + \binom{n}{3}8^3 + \binom{n}{4}8^4 + \dots + \binom{n}{n}8^n}{8^2}$ $\alpha = \binom{n}{2}8^{2-2} + \binom{n}{3}8^{3-2} + \binom{n}{4}8^{4-2} + \dots + \binom{n}{n}8^{n-2}$ $\alpha = \binom{n}{2} + \binom{n}{3}8 + \binom{n}{4}8^2 + \dots + \binom{n}{n}8^{n-2}$ This is our simplified expression for $\alpha$.

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Part 2: Deriving the expression for $\beta$}

Similarly, we are given the equation: $6^n - 5n - 1 = 25\beta$

Step 2.1: Express $6^n$ in binomial form We rewrite $6^n$ as $(1+5)^n$. This is analogous to the previous step, setting up the expression for binomial expansion with powers of 5, which will interact with the $25 = 5^2$ in the denominator. $6^n = (1+5)^n$

Step 2.2: Expand $(1+5)^n$ using the Binomial Theorem Using the expansion $(1+x)^n = 1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + x^n$ with $x=5$: $(1+5)^n = \binom{n}{0} + \binom{n}{1}(5) + \binom{n}{2}(5)^2 + \binom{n}{3}(5)^3 + \dots + \binom{n}{n}(5)^n$ Substituting $\binom{n}{0}=1$ and $\binom{n}{1}=n$: $(1+5)^n = 1 + 5n + \binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n$

Step 2.3: Substitute the expansion and simplify for $\beta$ Now, substitute this expanded form of $(1+5)^n$ back into the original equation for $\beta$: $25\beta = \left( 1 + 5n + \binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n \right) - 5n - 1$ Again, the terms $1$ and $5n$ from the expansion cancel out with the $-5n-1$ terms: $25\beta = (1-1) + (5n-5n) + \binom{n}{2}5^2 + \binom{n}{3}5^3 + \dots + \binom{n}{n}5^n$ $25\beta = \binom{n}{2}5^2 + \binom{n}{3}5^3 + \binom{n}{4}5^4 + \dots + \binom{n}{n}5^n$ Finally, divide both sides by $25$ (which is $5^2$) to find $\beta$. Each term in the summation is divided by $5^2$, reducing the power of 5 by two: $\beta = \frac{\binom{n}{2}5^2 + \binom{n}{3}5^3 + \binom{n}{4}5^4 + \dots + \binom{n}{n}5^n}{5^2}$ $\beta = \binom{n}{2}5^{2-2} + \binom{n}{3}5^{3-2} + \binom{n}{4}5^{4-2} + \dots + \binom{n}{n}5^{n-2}$ $\beta = \binom{n}{2} + \binom{n}{3}5 + \binom{n}{4}5^2 + \dots + \binom{n}{n}5^{n-2}$ This is our simplified expression for $\beta$.

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Part 3: Calculating $\alpha - \beta$

Now that we have simplified expressions for $\alpha$ and $\beta$, we can find their difference: $\alpha - \beta = \left( \binom{n}{2} + \binom{n}{3}8 + \binom{n}{4}8^2 + \dots + \binom{n}{n}8^{n-2} \right) - \left( \binom{n}{2} + \binom{n}{3}5 + \binom{n}{4}5^2 + \dots + \binom{n}{n}5^{n-2} \right)$

Step 3.1: Group common binomial coefficients and simplify We group the terms that share the same binomial coefficient $\binom{n}{k}$: $\alpha - \beta = \left(\binom{n}{2} - \binom{n}{2}\right) + \left(\binom{n}{3}8 - \binom{n}{3}5\right) + \left(\binom{n}{4}8^2 - \binom{n}{4}5^2\right) + \dots + \left(\binom{n}{n}8^{n-2} - \binom{n}{n}5^{n-2}\right)$ The first terms, $\binom{n}{2} - \binom{n}{2}$, cancel each other out, resulting in 0. Then, we factor out the common binomial coefficients from the remaining terms: $\alpha - \beta = 0 + \binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \dots + \binom{n}{n}(8^{n-2}-5^{n-2})$ $\alpha - \beta = \binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \dots + \binom{n}{n}(8^{n-2}-5^{n-2})$

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Conclusion

By systematically applying the Binomial Theorem and simplifying the resulting expressions, we find that: $\alpha - \beta = \binom{n}{3}(8-5) + \binom{n}{4}(8^2-5^2) + \dots + \binom{n}{n}(8^{n-2}-5^{n-2})$ This expression matches Option (C).

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Tips for JEE Aspirants / Common Mistakes

1. Master the Binomial Theorem: A strong understanding of $(a+b)^n$ and especially $(1+x)^n$ expansions is fundamental. Remember the values of $\binom{n}{0}=1$ and $\binom{n}{1}=n$.
2. Strategic Binomial Expansion: When you see expressions like $9^n$ or $6^n$ combined with linear terms ($8n$, $5n$) and a constant ($1$), immediately think of expanding them as $(1+8)^n$ or $(1+5)^n$. The goal is often to create terms that will cancel out with the linear and constant terms, leaving only higher-order terms divisible by the given constant ($64$ or $25$).
3. Careful Algebraic Manipulation: Pay close attention to signs when subtracting expressions. Ensure that all terms are correctly grouped and factored.
4. Index Management: Be precise with the exponents and the range of summation for binomial coefficients. A common mistake is to miscount the starting or ending index, or the power of $x$ in each term after division. In this problem, after dividing by $x^2$, the term $\binom{n}{k}x^k$ becomes $\binom{n}{k}x^{k-2}$.
5. Option Comparison: Always compare your derived solution carefully with the given options. Sometimes, the options might be presented in a slightly different form, or there might be an error in the question or the marked correct answer itself. Based on our rigorous derivation, the result matches Option (C), despite the question marking (A) as correct. It's crucial to trust your correct mathematical derivation.

Key Takeaway: This problem is a classic application of the Binomial Theorem to simplify complex algebraic expressions. The key insight is to recognize how to manipulate the base of the exponential terms ($9^n$ and $6^n$) into a binomial form $(1+x)^n$ such that the lower-order terms cancel out, allowing for clean division and summation.