In the expansion of ([3]2+1[3]3)^n, n ~N, if the ratio of 15^ th term from the beginning t…

Question

In the expansion of $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}ight)^n, n \in \mathrm{~N}$, if the ratio of $15^{	ext {th }}$ term from the beginning to the $15^{	ext {th }}$ term from the end is $\frac{1}{6}$, then the value of ${ }^n \mathrm{C}_3$ is

Accepted Answer

In the expansion of $(a+b)^n$

$15^{\text {th }}$ term from beginning: $T_{15}={ }^n C_{14} a^{n-14} b^{14}$
$15^{\text {th }}$ term from end: $T_{15}^{\prime}={ }^n C_{14} b^{n-14} a^{14}$

$$\begin{aligned}
& \therefore \quad \frac{T_{15}}{T_{15}^{\prime}}=\frac{1}{6} \
& \left(\frac{a}{b}\right)^{n-28}=\frac{1}{6} \
& \left(6^{\frac{1}{3}}\right)^{n-28}=6^{-1} \
& \Rightarrow \quad \frac{n-28}{3}=-1 \
& n=25 \
& \therefore \quad{ }^{25} C_3=2300
\end{aligned}$$

Solutions · 1