Question

Let $m$ and $n$ be the coefficients of the seventh and thirteenth terms respectively in the expansion of \[\left( \frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2x^{\frac{2}{3}}} \right)^{18}.\]Then \[\left(\frac{n}{m}\right)^{\frac{1}{3}}\]is:

• A. $\frac{4}{9}$
• B. $\frac{1}{9}$
• C. $\frac{1}{4}$
• D. $\frac{9}{4}$

Answer 1

The Correct Option is D

In the binomial expansion of $(a+b)^{18}$, the general term is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$.
Using the formula for the general term in the binomial expansion, we can find the seventh and thirteenth terms of the given expansion.

Seventh term:
$T_7 = \binom{18}{6} \left(\frac{1}{3x^{\frac{1}{3}}}\right)^{12} \left(\frac{1}{2x^{\frac{1}{3}}}\right)^6$

$m = \binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6$

Thirteenth term:
$T_{13} = \binom{18}{12} \left(\frac{1}{3x^{\frac{1}{3}}}\right)^6 \left(\frac{1}{2x^{\frac{1}{3}}}\right)^{12}$

$n = \binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}$

Now, we need to find $\left(\frac{n}{m}\right)^{\frac{1}{3}}$.
$\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{\binom{18}{12} \left(\frac{1}{3}\right)^6 \left(\frac{1}{2}\right)^{12}}{\binom{18}{6} \left(\frac{1}{3}\right)^{12} \left(\frac{1}{2}\right)^6}\right)^{\frac{1}{3}}$

Simplifying the expression, we get:
$\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{\binom{18}{12}}{\binom{18}{6}} \times \left(\frac{1}{3}\right)^{-6} \times \left(\frac{1}{2}\right)^6\right)^{\frac{1}{3}}$

Using the property of binomial coefficients $\binom{n}{r} = \binom{n}{n-r}$, we can simplify further:
$\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\frac{\binom{18}{6}}{\binom{18}{6}} \times \left(\frac{1}{3}\right)^{-6} \times \left(\frac{1}{2}\right)^6\right)^{\frac{1}{3}} = \left(\frac{1}{3^{-6}} \times \frac{1}{2^6}\right)^{\frac{1}{3}} = \left(3^6 \times 2^{-6}\right)^{\frac{1}{3}} = \left(\frac{3^2}{2^2}\right) = \frac{9}{4}$

Therefore, the correct answer is (4).