JEE Main 2022 (Online) 29th July Morning ShiftQuestion 1
Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n$, in the increasing powers of $\frac{1}{\sqrt[4]{3}}$ be $\sqrt[4]{6}:1$. If the sixth term from the beginning is $\frac{\alpha}{\sqrt[4]{3}}$, then $\alpha$ is equal to ________.
Binomial Theorem · General Term and Middle Term
Answer$84$
Solution
Concept. In increasing powers of $3^{-1/4}$ the general term is ${}^{n}C_k\,2^{\frac{n-k}{4}}\,3^{-\frac k4}$; the $5$th from the end is $k=n-4$. Formulas used. (1) $T_{k+1}={}^{n}C_k\,a^{n-k}b^k$ and ${}^{n}C_4={}^{n}C_{n-4}$Step 1 - The ratio. Fifth from the start ($k=4$) over fifth from the end ($k=n-4$): $$\frac{2^{\frac{n-4}{4}}\,3^{-1}}{2^{1}\,3^{-\frac{n-4}{4}}}=2^{\frac{n-8}{4}}\,3^{\frac{n-8}{4}}=6^{\frac{n-8}{4}}.$$ Step 2 - Solve. $6^{\frac{n-8}{4}}=6^{1/4}$ gives $n=9$. Step 3 - Sixth term ($k=5$). $${}^{9}C_5\,2^{1}\,3^{-5/4}=\frac{252}{3\cdot3^{1/4}}=\frac{84}{\sqrt[4]3}.$$ So $\alpha=84$. Answer: 84