Binomial Theorem Question 2
Question 2 - 2024 (01 Feb Shift 2)
Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{n}{m}\right)^{\frac{1}{3}}$ is :
(1) $\frac{4}{9}$
(2) $\frac{1}{9}$
(3) $\frac{1}{4}$
(4) $\frac{9}{4}$
Show Answer
Answer (4)
Solution
$\left(\frac{x^{\frac{1}{3}}}{3}+\frac{x^{\frac{-2}{3}}}{18}\right)^{18}$
$t _7={ }^{18} c _6\left(\frac{x^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{6}={ }^{18} c _6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^{6}}$
$t _{13}={ }^{18} c _{12}\left(\frac{x^{\frac{1}{3}}}{3}\right)^{6}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} c _{12} \frac{1}{(3)^{6}} \cdot \frac{1}{2^{12}} \cdot x^{-6}$
$m={ }^{18} c _6 \cdot 3^{-12} \cdot 2^{-6}: n={ }^{18} c _{12} \cdot 2^{-12} \cdot 3^{-6}$
$\left(\frac{n}{m}\right)^{\frac{1}{3}}=\left(\frac{2^{-12} \cdot 3^{-6}}{3^{-12} \cdot 2^{-6}}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^{2}=\frac{9}{4}$
