Q.If the seventh terms from the beginning and the end in the expansion of $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^{n}$ are equal, then $n$ equals ______ .

The key idea here is the Symmetry Property of Binomial Terms. For an expansion $(a+b)^n$, the $k$-th term from the beginning and the $k$-th term from the end have the same binomial coefficient, but the powers of $a$ and $b$ are swapped.

1. The $k$-th term from the beginning in $(A+B)^n$ is $T_k = \binom{n}{k-1} A^{n-(k-1)} B^{k-1}$.
2. The $k$-th term from the end in $(A+B)^n$ is $T'_k = \binom{n}{k-1} A^{k-1} B^{n-(k-1)}$.
3. Given that the seventh terms ($k=7$) from the beginning and end are equal: $T_7 = T'_7 \implies \binom{n}{6} \left(\sqrt[3]{2}\right)^{n-6} \left(\frac{1}{\sqrt[3]{3}}\right)^6 = \binom{n}{6} \left(\sqrt[3]{2}\right)^6 \left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}$.
4. Cancelling $\binom{n}{6}$ (since $n \ge 6$) and rearranging, we get: …

In the expansion of $(a+b)^n$, the terms exhibit a beautiful symmetry. The $r$-th term from the beginning, $T_r$, and the $r$-th term from the end, $T'_r$, are related. Specifically, the $r$-th term from the beginning has the form $\binom{n}{r-1} a^{n-(r-1)} b^{r-1}$. The $r$-th term from the end is equivalent to the $(n+1 - r + 1)$-th term from the beginning, which is the $(n-r+2)$-th term from the beginning.

Let's denote the general term from the beginning as $T_{k+1}$.

For the terms to be equal, both their binomial coefficients and their variable parts must be identical.

1. Identify the components of the binomial expression:

   The given expression is $\left(\sqrt[3]{2} + \dfrac{1}{\sqrt[3]{3}}\right)^{n}$.

   Here, $a = \sqrt[3]{2} = 2^{1/3}$ and $b = \dfrac{1}{\sqrt[3]{3}} = 3^{-1/3}$.

2. Find the seventh term from the beginning:

   For the seventh term from the beginning, we set $k+1 = 7$, so $k=6$.

   $T_7 = \binom{n}{6} (2^{1/3})^{n-6} (3^{-1/3})^6$

   $T_7 = \binom{n}{6} 2^{(n-6)/3} 3^{-6/3}$

   $T_7 = \binom{n}{6} 2^{(n-6)/3} 3^{-2}$

3. Find the seventh term from the end:

   The total number of terms in the expansion of $(a+b)^n$ is $n+1$.

   The $r$-th term from the end is the $(n+1 - r + 1)$-th term from the beginning.

   For the seventh term from the end, $r=7$.

   So, the seventh term from the end is the $(n+1 - 7 + 1)$-th term from the beginning, which simplifies to the $(n-5)$-th term from the beginning.

   Let's call this $T'_{7}$. This means we need to find $T_{k+1}$ where $k+1 = n-5$, so $k = n-6$.

   $T'_{7} = T_{n-5} = \binom{n}{n-6} (2^{1/3})^{n-(n-6)} (3^{-1/3})^{n-6}$

   $T'_{7} = \binom{n}{n-6} (2^{1/3})^6 (3^{-1/3})^{n-6}$

   $T'_{7} = \binom{n}{n-6} 2^{6/3} 3^{-(n-6)/3}$

   $T'_{7} = \binom{n}{n-6} 2^2 3^{-(n-6)/3}$

   Tip

   A quicker way to think about the $k$-th term from the end is to swap $a$ and $b$ in the original binomial and find the $k$-th term from the beginning. The expansion of $(b+a)^n$ has the same terms as $(a+b)^n)$, just in reverse order. So, the 7th term from the end of $(a+b)^n$ is the 7th term from the beginning of $(b+a)^n$.

   $T_7 \text{ of } (b+a)^n = \binom{n}{6} b^{n-6} a^6 = \binom{n}{6} (3^{-1/3})^{n-6} (2^{1/3})^6$.

   This matches our $T'_{7}$ because $\binom{n}{n-6} = \binom{n}{6}$.

4. Equate the seventh terms from the beginning and the end: …