JEE Mathematics 2024 Question with Solution

Given: We need the coefficients $m$ and $n$ of the seventh and thirteenth terms respectively in the expansion.

Find: The value of $\left(\frac{n}{m}\right)^{\frac{1}{3}}$.

For the binomial expansion, the general term is

$$T_{k+1} = \binom{18}{k}\left(\frac{1}{3}x^{\frac{1}{3}}\right)^{18-k}\left(\frac{1}{2x^{\frac{2}{3}}}\right)^k$$

So,

$$T_{k+1} = \binom{18}{k}\left(\frac{1}{3}\right)^{18-k}\left(\frac{1}{2}\right)^k x^{\frac{18-k}{3}}x^{-\frac{2k}{3}}$$

Hence the power of $x$ becomes

$$x^{\frac{18-k-2k}{3}} = x^{6-k}$$

The seventh term corresponds to $k=6$, so

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

The thirteenth term corresponds to $k=12$, so

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

Now,

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

Using $\binom{18}{12}=\binom{18}{6}$,

$$\left(\frac{n}{m}\right)^{\frac{1}{3}} = \left(\left(\frac{1}{3}\right)^{-6}\left(\frac{1}{2}\right)^6\right)^{\frac{1}{3}} = \left(\frac{3^6}{2^6}\right)^{\frac{1}{3}} = \frac{9}{4}$$

Therefore, the correct option is D.

The solution concludes the required value as $\frac{9}{4}$, matching option D.

Given: The seventh and thirteenth term coefficients are $m$ and $n$.

Find: The required ratio from the extracted coefficients.

Write the general term of the binomial part as

$$T_{r+1}=\binom{18}{r}a^{18-r}b^r$$

with

$$a=\frac{1}{3}x^{\frac{1}{3}}, \qquad b=\frac{1}{2x^{\frac{2}{3}}}$$

Then

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

so

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

Also,

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

so

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

Therefore,

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

Now use symmetry of binomial coefficients:

$$\binom{18}{12}=\binom{18}{6}$$

Hence,

$$\left(\frac{n}{m}\right)^{\frac{1}{3}}=\left(3^6\times 2^{-6}\right)^{\frac{1}{3}}=\left(\frac{3^6}{2^6}\right)^{\frac{1}{3}}=\frac{3^2}{2^2}=\frac{9}{4}$$

Therefore, the correct option is D.