Solution: Maximizing the Term Independent of x in Binomial Expansion

Key Concepts and Formulas

This problem combines two important mathematical concepts: the Binomial Theorem for finding specific terms in an expansion, and the Arithmetic Mean - Geometric Mean (AM-GM) Inequality for finding the maximum value of an expression.

1. Binomial Theorem (General Term): For a binomial expansion of the form $(A+B)^N$, the general term (or $(r+1)^{th}$ term) is given by: $T_{r+1} = \binom{N}{r} A^{N-r} B^r$ where $\binom{N}{r} = \frac{N!}{r!(N-r)!}$ is the binomial coefficient.

2. Term Independent of $x$: A term is considered "independent of $x$" if the variable $x$ does not appear in it. This means the exponent of $x$ in that term must be zero ($x^0 = 1$).

3. AM-GM Inequality: For any set of $n$ non-negative real numbers $a_1, a_2, \dots, a_n$, the arithmetic mean is always greater than or equal to their geometric mean: $\frac{a_1 + a_2 + \dots + a_n}{n} \ge \sqrt[n]{a_1 a_2 \dots a_n}$ Equality holds if and only if all the numbers are equal ($a_1 = a_2 = \dots = a_n$). This inequality is frequently used to find the maximum or minimum values of expressions.

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Step-by-Step Working with Explanations

Step 1: Determine the General Term of the Expansion

We are given the binomial expression ${\left( {\alpha {x^{{1 \over 9}}} + \beta {x^{ - {1 \over 6}}}} \right)^{10}}$. Here, $A = \alpha x^{1/9}$, $B = \beta x^{-1/6}$, and $N = 10$.

Applying the general term formula $T_{r+1} = \binom{N}{r} A^{N-r} B^r$: $T_{r+1} = \binom{10}{r} \left(\alpha x^{1/9}\right)^{10-r} \left(\beta x^{-1/6}\right)^r$

Now, we separate the coefficients and the powers of $x$: $T_{r+1} = \binom{10}{r} \alpha^{10-r} (x^{1/9})^{10-r} \beta^r (x^{-1/6})^r$ Using the exponent rule $(a^m)^n = a^{mn}$: $T_{r+1} = \binom{10}{r} \alpha^{10-r} \beta^r x^{\frac{10-r}{9}} x^{-\frac{r}{6}}$ Combining the powers of $x$ using $x^m x^n = x^{m+n}$: $T_{r+1} = \binom{10}{r} \alpha^{10-r} \beta^r x^{\frac{10-r}{9} - \frac{r}{6}}$

Explanation: We start by writing the general term because we need to identify the structure of each term in the expansion. This allows us to isolate the part containing $x$ and determine when it becomes independent of $x$.

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Step 2: Find the Value of $r$ for the Term Independent of $x$

For a term to be independent of $x$, the exponent of $x$ must be zero. So, we set the exponent of $x$ from the general term to zero: $\frac{10-r}{9} - \frac{r}{6} = 0$

To solve for $r$, find a common denominator for $9$ and $6$, which is $18$. Multiply the entire equation by $18$: $18 \left( \frac{10-r}{9} \right) - 18 \left( \frac{r}{6} \right) = 0 \times 18$ $2(10-r) - 3r = 0$ $20 - 2r - 3r = 0$ $20 - 5r = 0$ $5r = 20$ $r = 4$

Explanation: Setting the exponent of $x$ to zero is the direct mathematical translation of the condition "term independent of $x$". Solving for $r$ tells us which specific term in the expansion satisfies this condition. The value of $r$ must be a non-negative integer and less than or equal to $N$ (here, $0 \le r \le 10$), which $r=4$ fulfills.

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Step 3: Calculate the Term Independent of $x$

Now that we know $r=4$, we substitute this value back into our general term expression: $T_{r+1} = T_{4+1} = T_5$ $T_5 = \binom{10}{4} \alpha^{10-4} \beta^4 x^{\frac{10-4}{9} - \frac{4}{6}}$ Since we already found that the exponent of $x$ is $0$ when $r=4$, the $x$ term becomes $x^0 = 1$. So, the term independent of $x$ is: $T_5 = \binom{10}{4} \alpha^6 \beta^4$

First, calculate the binomial coefficient $\binom{10}{4}$: $\binom{10}{4} = \frac{10!}{4!(10-4)!} = \frac{10!}{4!6!} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 10 \times 3 \times 7 = 210$ Therefore, the term independent of $x$ is $210 \alpha^6 \beta^4$.

Explanation: We substitute $r=4$ to get the concrete form of the term we are interested in. This term now depends only on $\alpha$ and $\beta$, whose values are constrained by the given equation.

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Step 4: Maximize the Term using AM-GM Inequality

We need to find the maximum value of $210 \alpha^6 \beta^4$, given the condition $\alpha^3 + \beta^2 = 4$ and $\alpha > 0, \beta > 0$. The expression to maximize is $P = \alpha^6 \beta^4 = (\alpha^3)^2 (\beta^2)^2$. Notice that the terms in the given sum ($\alpha^3$ and $\beta^2$) correspond directly to the bases in the expression we want to maximize. This is a strong indicator to use AM-GM inequality.

Consider the two positive terms $\alpha^3$ and $\beta^2$. According to AM-GM inequality for two terms: $\frac{\alpha^3 + \beta^2}{2} \ge \sqrt{\alpha^3 \beta^2}$

We are given $\alpha^3 + \beta^2 = 4$. Substitute this into the inequality: $\frac{4}{2} \ge \sqrt{\alpha^3 \beta^2}$ $2 \ge \sqrt{\alpha^3 \beta^2}$

To get rid of the square root and match the powers needed for $\alpha^6 \beta^4$, we square both sides of the inequality: $(2)^2 \ge (\sqrt{\alpha^3 \beta^2})^2$ $4 \ge \alpha^3 \beta^2$

Now, to obtain $\alpha^6 \beta^4$, we need to square both sides again: $(4)^2 \ge (\alpha^3 \beta^2)^2$ $16 \ge \alpha^6 \beta^4$

This inequality tells us that the maximum possible value for $\alpha^6 \beta^4$ is $16$. The equality in AM-GM holds when the terms are equal, i.e., $\alpha^3 = \beta^2$. Substituting this into $\alpha^3 + \beta^2 = 4$: $\alpha^3 + \alpha^3 = 4 \Rightarrow 2\alpha^3 = 4 \Rightarrow \alpha^3 = 2$. Then $\beta^2 = 2$. Since $\alpha, \beta > 0$, this is a valid scenario where the maximum value is achieved.

Explanation: The AM-GM inequality is powerful for finding maximums. We strategically applied it to the terms $\alpha^3$ and $\beta^2$ because they were related by a sum and we needed to maximize their product (raised to certain powers). The maximum value occurs when the terms in the AM-GM are equal.

Common Mistake: A common pitfall is incorrectly identifying the terms to apply AM-GM to, especially when the powers don't align directly. Here, the structure $(\alpha^3)^2 (\beta^2)^2$ perfectly matched the sum $\alpha^3 + \beta^2$. If the powers were more complex, one might need to use weighted AM-GM or a different strategy.

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Step 5: Calculate the Maximum Value of the Term and Find $k$

The term independent of $x$ was found to be $T_5 = 210 \alpha^6 \beta^4$. We found the maximum value of $\alpha^6 \beta^4$ is $16$. Therefore, the maximum value of the term independent of $x$ is: $T_{5, \text{max}} = 210 \times 16$ $T_{5, \text{max}} = 3360$

The problem states that the maximum value of the term independent of $x$ is $10k$. So, we have: $10k = 3360$ Divide by $10$ to find $k$: $k = \frac{3360}{10}$ $k = 336$

Explanation: This is the final calculation step, where we combine all our intermediate results to arrive at the solution for $k$ as requested by the problem.

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Summary and Key Takeaway

This problem effectively tests your understanding of two core mathematical tools: the Binomial Theorem and the AM-GM Inequality. The process involved:

1. Formulating the general term of the binomial expansion.
2. Using the property of "term independent of $x$" (exponent of $x$ is zero) to find the specific term.
3. Applying the AM-GM inequality to maximize the algebraic expression involving $\alpha$ and $\beta$, given their sum.
4. Finally, calculating the unknown constant $k$.

Key Takeaway: For maximization/minimization problems involving sums and products of non-negative terms, always consider the AM-GM inequality. Pay close attention to the powers of variables to strategically choose the terms for AM-GM application.