Q.Find $n$ in the binomial $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^{n}$ if the ratio of $7^{\text{th}}$ term from the beginning to the $7^{\text{th}}$ term from the end is $\dfrac{1}{6}$.

Concept: Symmetry property of binomial coefficients and the general term formula.

In the expansion of $(a + b)^n$, the $r^{\text{th}}$ term from the beginning is $T_r = \binom{n}{r-1} a^{n-r+1} b^{r-1}$, while the $r^{\text{th}}$ term from the end is $T_{n-r+2} = \binom{n}{n-r+1} a^{r-1} b^{n-r+1}$.

For our binomial with $a = \sqrt[3]{2} = 2^{1/3}$ and $b = \frac{1}{\sqrt[3]{3}} = 3^{-1/3}$:

The $7^{\text{th}}$ term from the beginning is:

The $7^{\text{th}}$ term from the end is:

Taking their ratio:

This gives us $2^{(n-12)/3} \cdot 3^{(12-n)/3} = \frac{1}{6}$, which simplifies to:

$\left(\frac{2}{3}\right)^{(n-12)/3} = \frac{1}{6}$ …

The key insight here rests on understanding how terms in a binomial expansion relate when counted from opposite ends. In $(a + b)^n$, the $r^{\text{th}}$ term from the beginning is $T_r = \binom{n}{r-1} a^{n-r+1} b^{r-1}$, while the $r^{\text{th}}$ term from the end is $T_{n-r+2} = \binom{n}{n-r+1} a^{r-1} b^{n-r+1}$. Notice that the binomial coefficients are equal by symmetry: $\binom{n}{r-1} = \binom{n}{n-r+1}$. What changes is the distribution of powers between $a$ and $b$.

Let me denote $a = \sqrt[3]{2} = 2^{1/3}$ and $b = \frac{1}{\sqrt[3]{3}} = 3^{-1/3}$ for clarity.

1. Write the $7^{\text{th}}$ term from the beginning

The general term in the expansion is:

For the $7^{\text{th}}$ term, we have $k = 6$:

2. Write the $7^{\text{th}}$ term from the end

The $7^{\text{th}}$ term from the end is the $(n - 6 + 1)^{\text{th}} = (n-5)^{\text{th}}$ term from the beginning. This corresponds to $k = n - 6$:

Since $\binom{n}{n-6} = \binom{n}{6}$, we have:

3. Set up the ratio

We're told that:

Substituting our expressions:

The binomial coefficients cancel: …