Understanding the Binomial Expansion and General Term

This problem requires us to find the ratio of specific terms within a binomial expansion. The core concept is the general term formula for a binomial expansion.

Key Concept: The general term, often denoted as $T_{r+1}$, in the binomial expansion of $(a+b)^n$ is given by the formula: $T_{r+1} = {}^{n}C_r a^{n-r} b^r$ where:

• $n$ is the power of the binomial.
• $r$ is the index, starting from $0$.
• ${}^{n}C_r = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

For calculating a term from the end, there are two standard approaches:

1. The $k^{th}$ term from the end of $(a+b)^n$ is equivalent to the $(n-k+2)^{th}$ term from the beginning.
2. Alternatively, the $k^{th}$ term from the end of $(a+b)^n$ is the same as the $k^{th}$ term from the beginning of $(b+a)^n$. This approach often simplifies ratio calculations, especially when the binomial coefficients are involved.

Step-by-Step Solution

1. Identify the Components of the Binomial Expansion First, let's extract the necessary components from the given binomial expansion: ${\left( {{2^{1/3}} + {1 \over {2{{\left( 3 \right)}^{1/3}}}}} \right)^{10}}$ Here, we have:

• The first term, $a = 2^{1/3}$
• The second term, $b = \frac{1}{2 \cdot 3^{1/3}}$
• The power of the binomial, $n = 10$

2. Calculate the 5th Term from the Beginning ($T_5$) To find the 5th term, we use the general term formula $T_{r+1} = {}^{n}C_r a^{n-r} b^r$. Since we want the 5th term ($T_5$), we set $r+1 = 5$, which means $r=4$.

Now, substitute the values of $n=10$, $r=4$, $a=2^{1/3}$, and $b=\frac{1}{2 \cdot 3^{1/3}}$ into the formula: $T_5 = {}^{10}C_4 \left(2^{1/3}\right)^{10-4} \left(\frac{1}{2 \cdot 3^{1/3}}\right)^4$ $T_5 = {}^{10}C_4 \left(2^{1/3}\right)^6 \left(\frac{1}{2 \cdot 3^{1/3}}\right)^4$

Next, we simplify the powers using the exponent rules $(x^m)^n = x^{mn}$ and $(\frac{X}{Y})^n = \frac{X^n}{Y^n}$:

• $\left(2^{1/3}\right)^6 = 2^{(1/3) \cdot 6} = 2^2 = 4$
• $\left(\frac{1}{2 \cdot 3^{1/3}}\right)^4 = \frac{1^4}{2^4 \cdot (3^{1/3})^4} = \frac{1}{16 \cdot 3^{4/3}}$

Substitute these simplified terms back into the expression for $T_5$: $T_5 = {}^{10}C_4 \cdot 4 \cdot \frac{1}{16 \cdot 3^{4/3}}$ $T_5 = {}^{10}C_4 \cdot \frac{1}{4 \cdot 3^{4/3}}$

3. Calculate the 5th Term from the End ($T'_5$) We'll use the property that the $k^{th}$ term from the end of $(a+b)^n$ is the same as the $k^{th}$ term from the beginning of $(b+a)^n$. This means we effectively swap $a$ and $b$ for the term from the end.

So, to find the 5th term from the end, we consider the expansion of $\left(\frac{1}{2 \cdot 3^{1/3}} + 2^{1/3}\right)^{10}$. In this "reversed" binomial:

• The new first term, $a' = \frac{1}{2 \cdot 3^{1/3}}$
• The new second term, $b' = 2^{1/3}$
• The power of the binomial remains $n = 10$

For the 5th term from the beginning of this reversed expansion ($T'_5$), we again set $r=4$: $T'_5 = {}^{10}C_4 \left(\frac{1}{2 \cdot 3^{1/3}}\right)^{10-4} \left(2^{1/3}\right)^4$ $T'_5 = {}^{10}C_4 \left(\frac{1}{2 \cdot 3^{1/3}}\right)^6 \left(2^{1/3}\right)^4$

Now, simplify the powers:

• $\left(\frac{1}{2 \cdot 3^{1/3}}\right)^6 = \frac{1^6}{2^6 \cdot (3^{1/3})^6} = \frac{1}{64 \cdot 3^2} = \frac{1}{64 \cdot 9} = \frac{1}{576}$
• $\left(2^{1/3}\right)^4 = 2^{4/3}$

Substitute these simplified terms back into the expression for $T'_5$: $T'_5 = {}^{10}C_4 \cdot \frac{1}{576} \cdot 2^{4/3}$ $T'_5 = {}^{10}C_4 \cdot \frac{2^{4/3}}{576}$

4. Form the Ratio $\frac{T_5}{T'_5}$ Now we compute the ratio of the 5th term from the beginning ($T_5$) to the 5th term from the end ($T'_5$): $\frac{T_5}{T'_5} = \frac{{}^{10}C_4 \cdot \frac{1}{4 \cdot 3^{4/3}}}{{}^{10}C_4 \cdot \frac{2^{4/3}}{576}}$ The binomial coefficient ${}^{10}C_4$ is common to both terms and cancels out: $\frac{T_5}{T'_5} = \frac{\frac{1}{4 \cdot 3^{4/3}}}{\frac{2^{4/3}}{576}}$ To simplify this complex fraction, we multiply the numerator by the reciprocal of the denominator: $\frac{T_5}{T'_5} = \frac{1}{4 \cdot 3^{4/3}} \cdot \frac{576}{2^{4/3}}$ $\frac{T_5}{T'_5} = \frac{576}{4 \cdot 3^{4/3} \cdot 2^{4/3}}$ Now, simplify the numerical and exponential terms:

• Divide $576$ by $4$: $576 \div 4 = 144$.
• Group terms with the same fractional exponent in the denominator: $3^{4/3} \cdot 2^{4/3} = (3 \cdot 2)^{4/3} = 6^{4/3}$.

So, the ratio becomes: $\frac{T_5}{T'_5} = \frac{144}{6^{4/3}}$ To simplify further and match the options, we can rewrite $6^{4/3}$ as $6^{1 + 1/3} = 6^1 \cdot 6^{1/3} = 6 \cdot 6^{1/3}$. $\frac{T_5}{T'_5} = \frac{144}{6 \cdot 6^{1/3}}$ Divide $144$ by $6$: $144 \div 6 = 24$. $\frac{T_5}{T'_5} = \frac{24}{6^{1/3}}$ To express this in a form that includes $6^{2/3}$ or $36^{1/3}$, we can write $24$ as $4 \cdot 6$: $\frac{T_5}{T'_5} = \frac{4 \cdot 6}{6^{1/3}}$ Using the exponent rule $\frac{x^m}{x^n} = x^{m-n}$: $\frac{T_5}{T'_5} = 4 \cdot 6^{1 - 1/3} = 4 \cdot 6^{2/3}$ Finally, we can rewrite $6^{2/3}$ as $(6^2)^{1/3} = 36^{1/3}$. $\frac{T_5}{T'_5} = 4 \cdot (36)^{1/3}$ This means the ratio of the 5th term from the beginning to the 5th term from the end is $4 \cdot (36)^{1/3} : 1$.

Important Note on Discrepancy: The derived mathematical solution, which is consistent with the provided "Current Solution," leads to $4 \cdot (36)^{1/3} : 1$. This matches Option (D). However, the problem statement indicates Option (A) as the correct answer. Based on the standard application of the binomial theorem and careful calculation, Option (D) is the mathematically correct outcome.

Tips for Binomial Theorem Problems

• Careful with 'r': Always remember that for the $k^{th}$ term ($T_k$), the value of $r$ in the general term formula $T_{r+1}$ is $k-1$.
• Terms from the End: For ratios involving terms from the end, the trick of considering the expansion of $(b+a)^n$ is often more efficient than calculating the equivalent term number from the beginning. This works particularly well because it often leads to cancellation of binomial coefficients.
• Master Exponent Rules: Mistakes in simplifying powers (especially fractional and negative exponents) are common. Always double-check each application of exponent rules like $x^m \cdot x^n = x^{m+n}$, $\frac{x^m}{x^n} = x^{m-n}$, and $(x^m)^n = x^{mn}$.
• Systematic Simplification: Group terms with common bases and simplify powers methodically to avoid errors and ensure the final answer matches the format of the options.

Summary and Key Takeaway

This problem is an excellent exercise in applying the general term formula of a binomial expansion and handling terms from the end. By effectively using the property that the $k^{th}$ term from the end of $(a+b)^n$ is the $k^{th}$ term from the beginning of $(b+a)^n$, we were able to set up and simplify the ratio efficiently. The key takeaway is the importance of precise application of the binomial term formula and diligent simplification using exponent rules to arrive at the correct form of the answer. The final ratio determined was $4 \cdot (36)^{1/3} : 1$.