Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of (\frac13 x^\frac13+\frac12 x^\frac23)^18 . Then (\fracnm)^\frac13 is :

Binomial Theorem PYQ — JEE Main 2024 · 1 Feb · Shift 2 · Q49 with Solution

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Correct answer: D

General term of the expansion

For $\left(\frac{1}{3}x^{1/3}+\frac{1}{2x^{2/3}}\right)^{18},$ the general term is $T_{r+1}=\binom{18}{r}\left(\frac{1}{3}x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{2/3}}\right)^r.$

Simplifying,

\binom{18}{r}\frac{1}{3^{18-r}2^r}x^{\frac{18-3r}{3}}.$$ So, $$T_{r+1}=\binom{18}{r}\frac{1}{3^{18-r}2^r}x^{6-r}.$$ The **coefficient** of the $(r+1)$-th term is therefore $$\binom{18}{r}\frac{1}{3^{18-r}2^r}.$$ --- 2. **Seventh term** Seventh term means $r+1=7 \Rightarrow r=6$. Thus, $$m=\binom{18}{6}\frac{1}{3^{12}2^6}.$$ --- 3. **Thirteenth term** Thirteenth term means $r+1=13 \Rightarrow r=12$. Thus, $$n=\binom{18}{12}\frac{1}{3^6 2^{12}}.$$ Using symmetry of binomial coefficients, $$\binom{18}{12}=\binom{18}{6}.$$ Hence, $$n=\binom{18}{6}\frac{1}{3^6 2^{12}}.$$ --- 4. **Compute $\dfrac{n}{m}$** $$\frac{n}{m}= rac{\binom{18}{6}\dfrac{1}{3^6 2^{12}}}{\binom{18}{6}\dfrac{1}{3^{12}2^6}}.$$ Canceling $\binom{18}{6}$, $$\frac{n}{m}=\frac{3^{12}2^6}{3^6 2^{12}}=\frac{3^6}{2^6}=\left(\frac{3}{2}\right)^6.$$ Therefore, $$\left(\frac{n}{m}\right)^{1/3}=\left(\left(\frac{3}{2}\right)^6\right)^{1/3}=\left(\frac{3}{2}\right)^2=\frac{9}{4}.$$ --- 5. **Check options** The value is $$\frac{9}{4},$$ which corresponds to **Option D**. --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So, the answer agrees with the stored answer.

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View written solutionFree

Correct answer: D

General term of the expansion

For $\left(\frac{1}{3}x^{1/3}+\frac{1}{2x^{2/3}}\right)^{18},$ the general term is $T_{r+1}=\binom{18}{r}\left(\frac{1}{3}x^{1/3}\right)^{18-r}\left(\frac{1}{2x^{2/3}}\right)^r.$

Simplifying,

\binom{18}{r}\frac{1}{3^{18-r}2^r}x^{\frac{18-3r}{3}}.$$ So, $$T_{r+1}=\binom{18}{r}\frac{1}{3^{18-r}2^r}x^{6-r}.$$ The **coefficient** of the $(r+1)$-th term is therefore $$\binom{18}{r}\frac{1}{3^{18-r}2^r}.$$ --- 2. **Seventh term** Seventh term means $r+1=7 \Rightarrow r=6$. Thus, $$m=\binom{18}{6}\frac{1}{3^{12}2^6}.$$ --- 3. **Thirteenth term** Thirteenth term means $r+1=13 \Rightarrow r=12$. Thus, $$n=\binom{18}{12}\frac{1}{3^6 2^{12}}.$$ Using symmetry of binomial coefficients, $$\binom{18}{12}=\binom{18}{6}.$$ Hence, $$n=\binom{18}{6}\frac{1}{3^6 2^{12}}.$$ --- 4. **Compute $\dfrac{n}{m}$** $$\frac{n}{m}= rac{\binom{18}{6}\dfrac{1}{3^6 2^{12}}}{\binom{18}{6}\dfrac{1}{3^{12}2^6}}.$$ Canceling $\binom{18}{6}$, $$\frac{n}{m}=\frac{3^{12}2^6}{3^6 2^{12}}=\frac{3^6}{2^6}=\left(\frac{3}{2}\right)^6.$$ Therefore, $$\left(\frac{n}{m}\right)^{1/3}=\left(\left(\frac{3}{2}\right)^6\right)^{1/3}=\left(\frac{3}{2}\right)^2=\frac{9}{4}.$$ --- 5. **Check options** The value is $$\frac{9}{4},$$ which corresponds to **Option D**. --- 6. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So, the answer agrees with the stored answer.

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