Introduction: Key Concepts

This problem utilizes the Binomial Theorem to find specific terms in an expansion. The core formula we will use is the general term in the binomial expansion of $(a+b)^n$, which gives the $(r+1)^{th}$ term:

$T_{r+1} = {}^n C_r a^{n-r} b^r$

where ${}^n C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

Another important concept is finding terms from the end of an expansion. For an expansion of $(a+b)^n$ which has $(n+1)$ terms, the $k^{th}$ term from the end is equivalent to the $(n-k+2)^{th}$ term from the beginning. This conversion simplifies calculations significantly.

We will also use fundamental exponent rules ($x^p \cdot x^q = x^{p+q}$, $x^p / x^q = x^{p-q}$, $(x^p)^q = x^{pq}$, $(xy)^p = x^p y^p$) and the symmetry property of binomial coefficients (${}^n C_r = {}^n C_{n-r}$).

Step 1: Expressing the General Term for the Given Expansion

The given binomial expansion is $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}$. To apply the general term formula, we first express the terms in the form $a=X^{p}$ and $b=Y^{q}$:

• $a = \sqrt[4]{2} = 2^{1/4}$
• $b = \frac{1}{\sqrt[4]{3}} = \frac{1}{3^{1/4}} = 3^{-1/4}$

Now, substitute these into the general term formula $T_{r+1} = {}^n C_r a^{n-r} b^r$: $T_{r+1} = {}^n C_r \left(2^{1/4}\right)^{n-r} \left(3^{-1/4}\right)^r$ Applying the exponent rule $(X^p)^q = X^{pq}$: $T_{r+1} = {}^n C_r 2^{\frac{n-r}{4}} 3^{-\frac{r}{4}}$ This formula will allow us to find any term in the expansion.

Step 2: Finding the Fifth Term from the Beginning ($T_5$)

To find the fifth term from the beginning, we set $r+1 = 5$. This implies $r=4$. Substitute $r=4$ into the general term formula derived in Step 1: $T_5 = {}^n C_4 2^{\frac{n-4}{4}} 3^{-\frac{4}{4}}$ Simplifying the exponent of 3: $T_5 = {}^n C_4 2^{\frac{n-4}{4}} 3^{-1}$ This is the expression for the fifth term from the beginning.

Step 3: Finding the Fifth Term from the End ($T'_5$)

The binomial expansion $(a+b)^n$ has $(n+1)$ terms. The $k^{th}$ term from the end is the $( (n+1) - k + 1 )^{th}$ term from the beginning. For the fifth term from the end, $k=5$. So, it is the $((n+1) - 5 + 1)^{th}$ term from the beginning, which simplifies to the $(n-3)^{th}$ term from the beginning.

To find the $(n-3)^{th}$ term from the beginning, we set $r+1 = n-3$, which means $r = n-4$. Substitute $r = n-4$ into the general term formula from Step 1: $T'_5 = T_{n-3} = {}^n C_{n-4} 2^{\frac{n-(n-4)}{4}} 3^{-\frac{n-4}{4}}$ Simplifying the exponent of 2: $T'_5 = {}^n C_{n-4} 2^{\frac{4}{4}} 3^{-\frac{n-4}{4}}$ $T'_5 = {}^n C_{n-4} 2^1 3^{-\frac{n-4}{4}}$ This is the expression for the fifth term from the end.

Tip: Be careful when converting terms from the end. A common mistake is to simply use $k$ instead of $(n-k+2)$ for the term number from the beginning. Also, remember the symmetry property ${}^n C_r = {}^n C_{n-r}$, which will be useful for simplification.

Step 4: Setting up the Ratio Equation

The problem states that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt[4]{6}:1$. This can be written as: $\frac{T_5}{T'_5} = \frac{\sqrt[4]{6}}{1} = 6^{1/4}$ Now, substitute the expressions for $T_5$ (from Step 2) and $T'_5$ (from Step 3) into this ratio: $\frac{{}^n C_4 2^{\frac{n-4}{4}} 3^{-1}}{{}^n C_{n-4} 2^1 3^{-\frac{n-4}{4}}} = 6^{1/4}$

Step 5: Solving for 'n'

We simplify the ratio equation. Recall the property of binomial coefficients: ${}^n C_r = {}^n C_{n-r}$. Therefore, ${}^n C_4 = {}^n C_{n-4}$. These terms cancel out in the ratio, which is a key simplification.

The equation now becomes: $\frac{2^{\frac{n-4}{4}} 3^{-1}}{2^1 3^{-\frac{n-4}{4}}} = 6^{1/4}$ Now, apply the exponent rule $X^p / X^q = X^{p-q}$ for terms with the same base (2 and 3): $2^{\left(\frac{n-4}{4}\right) - 1} \cdot 3^{(-1) - \left(-\frac{n-4}{4}\right)} = 6^{1/4}$ Simplify the exponents: For base 2: $\frac{n-4}{4} - 1 = \frac{n-4-4}{4} = \frac{n-8}{4}$ For base 3: $-1 + \frac{n-4}{4} = \frac{-4+n-4}{4} = \frac{n-8}{4}$

Substitute these simplified exponents back into the equation: $2^{\frac{n-8}{4}} \cdot 3^{\frac{n-8}{4}} = 6^{1/4}$ Now, use the exponent rule $(XY)^p = X^p Y^p$ in reverse: $(2 \cdot 3)^{\frac{n-8}{4}} = 6^{1/4}$ $6^{\frac{n-8}{4}} = 6^{1/4}$ Since the bases are equal, their exponents must also be equal: $\frac{n-8}{4} = \frac{1}{4}$ Multiply both sides by 4: $n-8 = 1$ $n = 9$ Thus, the value of $n$ is 9.

Step 6: Calculating the Sixth Term from the Beginning ($T_6$)

Now that we have $n=9$, we can find the sixth term from the beginning. For the sixth term, $r+1 = 6$, so $r=5$. Using the general term formula $T_{r+1} = {}^n C_r 2^{\frac{n-r}{4}} 3^{-\frac{r}{4}}$ with $n=9$ and $r=5$: $T_6 = {}^9 C_5 2^{\frac{9-5}{4}} 3^{-\frac{5}{4}}$ $T_6 = {}^9 C_5 2^{\frac{4}{4}} 3^{-\frac{5}{4}}$ $T_6 = {}^9 C_5 \cdot 2^1 \cdot 3^{-\frac{5}{4}}$ First, calculate the binomial coefficient ${}^9 C_5$: ${}^9 C_5 = \frac{9!}{5!(9-5)!} = \frac{9!}{5!4!} = \frac{9 \times 8 \times 7 \times 6 \times 5!}{5! \times 4 \times 3 \times 2 \times 1} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} = 9 \times 2 \times 7 = 126$ Substitute the value of ${}^9 C_5$ back into the expression for $T_6$: $T_6 = 126 \cdot 2 \cdot 3^{-\frac{5}{4}}$ $T_6 = 252 \cdot 3^{-\frac{5}{4}}$ We can rewrite $3^{-\frac{5}{4}}$ as $\frac{1}{3^{5/4}}$. Further, $3^{5/4} = 3^{1 + 1/4} = 3^1 \cdot 3^{1/4} = 3\sqrt[4]{3}$. So, $3^{-\frac{5}{4}} = \frac{1}{3\sqrt[4]{3}}$. $T_6 = 252 \cdot \frac{1}{3\sqrt[4]{3}}$ $T_6 = \frac{252}{3\sqrt[4]{3}}$ $T_6 = \frac{84}{\sqrt[4]{3}}$

Step 7: Determining 'alpha'

The problem states that the sixth term from the beginning is $\frac{\alpha}{\sqrt[4]{3}}$. From our calculation in Step 6, we found $T_6 = \frac{84}{\sqrt[4]{3}}$. Comparing these two expressions: $\frac{\alpha}{\sqrt[4]{3}} = \frac{84}{\sqrt[4]{3}}$ By direct comparison, we can see that $\alpha = 84$.

Tips and Common Mistakes

• Careful with Exponents: Errors often occur when manipulating fractional or negative exponents. Double-check each step of simplification.
• Order of Operations: Ensure you perform calculations in the correct order, especially when dealing with binomial coefficients and powers.
• Reading the Question: Pay close attention to whether a term is requested from the beginning or the end, and apply the correct formula for conversion if necessary.
• Simplification of Radicals: Rationalizing denominators is often expected, but in this case, keeping the $\sqrt[4]{3}$ in the denominator allowed for direct comparison with the given form $\frac{\alpha}{\sqrt[4]{3}}$.

Summary/Key Takeaway

This problem is a comprehensive test of binomial theorem applications. It requires accurate identification of terms (both from the beginning and the end), careful application of exponent laws, and the use of properties of binomial coefficients to simplify algebraic expressions. The sequential steps of determining $n$, then calculating the specific term, and finally equating it to the given form are crucial for arriving at the correct value of $\alpha$. Mastery of these concepts is essential for solving such problems efficiently.