JEE Main 2025 (Online) 4th April Morning ShiftQuestion 1
In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, $n\in\mathbb{N}$, if the ratio of $15^{\text{th}}$ term from the beginning to the $15^{\text{th}}$ term from the end is $\frac{1}{6}$, then the value of ${}^{n}C_3$ is
Binomial Theorem · General Term and Middle Term
Solution
Concept. The $k$-th term from the end of $(a+b)^n$ is the $k$-th term from the beginning of $(b+a)^n$. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$(2) $k$-th term from the end $={}^{n}C_{k-1}\,a^{k-1}b^{n-k+1}$Step 1 - The two terms. With $a=2^{1/3}$, $b=3^{-1/3}$: $$T_{15}={}^{n}C_{14}\,a^{n-14}b^{14},\qquad T'_{15}={}^{n}C_{14}\,a^{14}b^{n-14}.$$ Step 2 - Ratio. $$\frac{T_{15}}{T'_{15}}=\left(\frac ab\right)^{n-28}=\left(2^{1/3}3^{1/3}\right)^{n-28}=6^{\frac{n-28}{3}}.$$ Step 3 - Solve. $6^{\frac{n-28}{3}}=6^{-1}$ gives $n-28=-3$, so $n=25$. Step 4 - Value. ${}^{25}C_3=\dfrac{25\cdot24\cdot23}{6}=2300$. Answer: (B)