Key Concept: General Term in Binomial Expansion

The general term, $T_{r+1}$, in the binomial expansion of $(x+a)^n$ is given by the formula: $T_{r+1} = {}^n C_r x^{n-r} a^r$ where ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

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Step 1: Identify the terms and calculate the fifth term from the beginning ($T_5$)

The given expansion is $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{\mathrm{n}}$. We can rewrite the terms using exponent notation: $x = 2^{1/4}$ and $a = 3^{-1/4}$.

To find the fifth term from the beginning, we set $r+1 = 5$, which means $r=4$. Substituting $x = 2^{1/4}$, $a = 3^{-1/4}$, and $r=4$ into the general term formula: $T_5 = {}^n C_4 (2^{1/4})^{n-4} (3^{-1/4})^4$ Explanation: We apply the general term formula directly for the $5^{th}$ term by setting $r=4$.

Now, simplify the powers: $T_5 = {}^n C_4 \left(2^{\frac{n-4}{4}}\right) \left(3^{-\frac{4}{4}}\right)$ $T_5 = {}^n C_4 \cdot 2^{\frac{n-4}{4}} \cdot 3^{-1}$

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Step 2: Calculate the fifth term from the end ($T'_5$)

The fifth term from the end in the expansion of $(x+a)^n$ is equivalent to the fifth term from the beginning in the expansion of $(a+x)^n$. So, for the fifth term from the end, we consider the expansion of $\left(\frac{1}{\sqrt[4]{3}} + \sqrt[4]{2}\right)^{\mathrm{n}}$. Here, $x' = 3^{-1/4}$ and $a' = 2^{1/4}$. For the fifth term, $r=4$. Substituting $x' = 3^{-1/4}$, $a' = 2^{1/4}$, and $r=4$ into the general term formula: $T'_5 = {}^n C_4 (3^{-1/4})^{n-4} (2^{1/4})^4$ Explanation: By swapping the terms in the binomial $(x+a)^n$ to $(a+x)^n$, the $k^{th}$ term from the end becomes the $k^{th}$ term from the beginning in the new expansion. This simplifies calculation as the binomial coefficients ${}^n C_r$ remain the same whether calculated from beginning or end using this technique (since ${}^n C_k = {}^n C_{n-k}$).

Now, simplify the powers: $T'_5 = {}^n C_4 \left(3^{-\frac{n-4}{4}}\right) \left(2^{\frac{4}{4}}\right)$ $T'_5 = {}^n C_4 \cdot 3^{-\frac{n-4}{4}} \cdot 2^1$

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Step 3: Use the given ratio to find the value of $n$

The problem states that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6}:1$. So, $\frac{T_5}{T'_5} = \frac{\sqrt{6}}{1} = \sqrt{6}$.

Substitute the expressions for $T_5$ and $T'_5$: $\frac{{}^n C_4 \cdot 2^{\frac{n-4}{4}} \cdot 3^{-1}}{{}^n C_4 \cdot 3^{-\frac{n-4}{4}} \cdot 2^1} = \sqrt{6}$ Explanation: We set up the equation based on the given ratio to solve for the unknown exponent $n$.

Cancel out the common term ${}^n C_4$ from the numerator and denominator: $\frac{2^{\frac{n-4}{4}} \cdot 3^{-1}}{3^{-\frac{n-4}{4}} \cdot 2^1} = \sqrt{6}$

Apply the exponent rules $a^m / a^p = a^{m-p}$: $2^{\left(\frac{n-4}{4} - 1\right)} \cdot 3^{\left(-1 - \left(-\frac{n-4}{4}\right)\right)} = \sqrt{6}$ $2^{\frac{n-4-4}{4}} \cdot 3^{\frac{-4 + n-4}{4}} = \sqrt{6}$ $2^{\frac{n-8}{4}} \cdot 3^{\frac{n-8}{4}} = \sqrt{6}$ Explanation: We combine the terms with the same base by subtracting their exponents.

Now, apply the exponent rule $a^m b^m = (ab)^m$: $(2 \cdot 3)^{\frac{n-8}{4}} = \sqrt{6}$ $6^{\frac{n-8}{4}} = 6^{1/2}$ Explanation: Since the bases of the powers are different but the exponents are the same, we can multiply the bases. We also express $\sqrt{6}$ as $6^{1/2}$ to facilitate comparison.

Equate the exponents since the bases are the same: $\frac{n-8}{4} = \frac{1}{2}$ Explanation: If $a^b = a^c$, then $b=c$ (for $a \ne 0, 1, -1$).

Solve for $n$: $n-8 = 4 \times \frac{1}{2}$ $n-8 = 2$ $n = 2+8$ $n = 10$

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Step 4: Calculate the third term from the beginning ($T_3$)

Now that we have found $n=10$, we can calculate the third term from the beginning for the original expansion $\left(2^{1/4} + 3^{-1/4}\right)^{10}$. For the third term, we set $r+1 = 3$, which means $r=2$. Substitute $n=10$, $x = 2^{1/4}$, $a = 3^{-1/4}$, and $r=2$ into the general term formula: $T_3 = {}^{10} C_2 (2^{1/4})^{10-2} (3^{-1/4})^2$ Explanation: We use the general term formula again, but this time with the determined value of $n=10$ to find the specific $3^{rd}$ term.

Calculate ${}^{10} C_2$: ${}^{10} C_2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 45$

Substitute this value back and simplify the powers: $T_3 = 45 \cdot (2^{1/4})^8 \cdot (3^{-1/4})^2$ $T_3 = 45 \cdot 2^{\frac{8}{4}} \cdot 3^{-\frac{2}{4}}$ $T_3 = 45 \cdot 2^2 \cdot 3^{-1/2}$ $T_3 = 45 \cdot 4 \cdot \frac{1}{\sqrt{3}}$ $T_3 = 180 \cdot \frac{1}{\sqrt{3}}$ Explanation: Simplify the binomial coefficient and the exponential terms. Remember that $a^{-m} = 1/a^m$.

To rationalize the denominator, multiply the numerator and denominator by $\sqrt{3}$: $T_3 = 180 \cdot \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$ $T_3 = \frac{180 \sqrt{3}}{3}$ $T_3 = 60 \sqrt{3}$ Explanation: Rationalizing the denominator removes the radical from the bottom of the fraction, presenting the answer in a standard form.

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Tips and Common Mistakes to Avoid:

1. Exponent Rules: Be very careful with exponent rules, especially when dealing with fractions and negative signs. A common mistake is misapplying $a^m/a^p$ or $(a^m)^p$.
2. Terms from the End: Remember the trick to find terms from the end: the $k^{th}$ term from the end of $(x+a)^n$ is the $k^{th}$ term from the beginning of $(a+x)^n$. Alternatively, the $k^{th}$ term from the end is the $(n-k+2)^{th}$ term from the beginning.
3. Binomial Coefficients: Ensure accurate calculation of ${}^n C_r$.
4. Rationalizing Denominators: Always rationalize denominators in the final answer if they contain radicals.

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Summary:

By systematically applying the general term formula for binomial expansion, we first set up expressions for the fifth term from the beginning and the fifth term from the end. Using the given ratio, we formed an equation that allowed us to solve for $n$, the power of the binomial. Once $n=10$ was determined, we used the general term formula again to calculate the third term from the beginning, simplifying it to its final rationalized form. The final answer is $60\sqrt{3}$.