Animated Solution for Mathematics - Binomial Theorem: A ratio of the 5th term from the beginning to the 5th term from the end in the binomial expansion of $(2^{1/3}+\frac{1}{2(3{)}^{1/3}}{)}^{10}$ is :

Welcome, future engineers! Today, we are going to unravel a problem that looks like a messy algebraic nightmare but is actually a masterclass in symmetry and pattern recognition.

We are looking at the binomial expansion of:

The goal is to find the ratio of the 5th term from the beginning to the 5th term from the end. We have a binomial $(a+b{)}^n$ where $n=10$, meaning there are $n+1=11$ terms in total.

Here, our first term is $a=2^{\frac{1}{3}}$ and our second term is $b=\frac{1}{2\cdot 3^{\frac{1}{3}}}$.

To find any term, we use the general term formula:

Here is where many students stumble: the index $r$. For the 5th term, $r=4$, as $r$ is always one less than the term number.

Substituting our values for $T_5$:

Simplifying the powers, we find $(2^{\frac{1}{3}}{)}^6=2^2=4$ and the second part becomes $\frac{1}{16\cdot 3^{\frac{4}{3}}}$. Thus:

Now, for the 5th term from the end, we use the property that the $k^{\text{th}}$ term from the end of $(a+b{)}^n$ is the $k^{\text{th}}$ term from the beginning of $(b+a{)}^n$. We simply swap the terms.

Now our first term is $\frac{1}{2\cdot 3^{\frac{1}{3}}}$ and our second is $2^{\frac{1}{3}}$. Calculating $T_5^{\prime }$ with $r=4$:

Simplifying this expression, we obtain:

Now, we set up the ratio $\frac{T_5}{T_5^{\prime }}$. The binomial coefficients $\left(\genfrac{}{}{0ex}{}{10}{4}\right)$ cancel out instantly:

Simplifying the constants, $\frac{64}{4}=16$, and $16\cdot 9=144$. The denominator becomes $3^{\frac{4}{3}}\cdot 2^{\frac{4}{3}}=(3\cdot 2{)}^{\frac{4}{3}}=6^{\frac{4}{3}}$.

We can write $6^{\frac{4}{3}}$ as $6\cdot 6^{\frac{1}{3}}$. Dividing $144$ by $6$ gives $24$:

Since $6^{\frac{2}{3}}=(36{)}^{\frac{1}{3}}$, the final ratio is: $4(36{)}^{\frac{1}{3}}:1$